Learn Coding | OnlineExaminer

Basics 6 Loops & Patterns 9 Digits 24 Primes & Divisors 8 Arrays 14 Strings 10 Matrices 4

Showing 75 of 75 problems · 0 learned

Topic 1 · 6 problems

Basics

Input, output, formulas and if-else. Your first programs.

1 Odd or Even Easy if-else modulus (%)

Write a program that takes an integer as input and prints "Odd" if the number is odd, otherwise "Even". The program should work for both positive and negative integers.

Sample input

Sample output

Odd

Tip: Test with n % 2 == 0, not n % 2 == 1. In Java, -3 % 2 is -1, so the second check would wrongly call -3 "Even".

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class OddOrEven {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter an integer: ");
        int n = sc.nextInt();

        // Even numbers leave no remainder when divided by 2
        if (n % 2 == 0) {
            System.out.println("Even");
        } else {
            System.out.println("Odd");
        }
    }
}

Solved by 2 students in Fun Program Battle

2 Smallest of Three Easy if-else comparison

Write a program to read three different integers and print the smallest of them. For example, if the input is 10, 20 and 5, the output should be 5.

Sample input

10 20 5

Sample output

Smallest number = 5

Tip: Java can also do this in one line with Math.min(Math.min(a, b), c), but writing the comparisons yourself is the better practice.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class SmallestOfThree {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter three numbers: ");
        int a = sc.nextInt();
        int b = sc.nextInt();
        int c = sc.nextInt();

        int smallest = a;          // assume the first number is the smallest
        if (b < smallest) {
            smallest = b;
        }
        if (c < smallest) {
            smallest = c;
        }
        System.out.println("Smallest number = " + smallest);
    }
}

Solved by 2 students in Fun Program Battle

3 Leap Year Easy logical operators if-else

Write a program to check whether a given year is a leap year. Print "Leap Year" if it is, otherwise print "Not a Leap Year". A year is a leap year if it is divisible by 4 but not by 100, or if it is divisible by 400.

Sample input

Sample output

Leap Year

Tip: 1900 is divisible by 4 but is NOT a leap year, because it is divisible by 100 and not by 400. 2000 is a leap year. Always test these two years.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class LeapYear {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a year: ");
        int y = sc.nextInt();

        // Divisible by 4 but not by 100, OR divisible by 400
        if ((y % 4 == 0 && y % 100 != 0) || (y % 400 == 0)) {
            System.out.println("Leap Year");
        } else {
            System.out.println("Not a Leap Year");
        }
    }
}

Solved by 3 students in Fun Program Battle

4 Taxi Fare Easy arithmetic double formatting

Write a program to calculate the total fare for a taxi ride. The base fare is $2.50, plus $0.25 per mile travelled and $0.50 per minute of travel time. Input the distance in miles and the time in minutes, then display the total fare.

Sample input

10
20

Sample output

Total fare = 15.00

Tip: Use double for money values and String.format("%.2f", value) to always show two decimal places.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class TaxiFare {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter distance in miles: ");
        double distance = sc.nextDouble();
        System.out.print("Enter time in minutes: ");
        double time = sc.nextDouble();

        double fare = 2.50 + 0.25 * distance + 0.50 * time;

        System.out.println("Total fare = " + String.format("%.2f", fare));
    }
}

Solved by 4 students in Fun Program Battle

5 Compound Interest Easy formula Math.pow double

Write a program to calculate the amount an investment grows to. Input the principal P, the annual rate of interest R (as a percentage) and the number of years N. Interest is compounded annually, so the amount is A = P x (1 + R/100)^N. Print the final amount.

Sample input

10000
10
2

Sample output

Amount after 2 years = 12100.00

Tip: Math.pow(x, y) returns x raised to the power y as a double.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class CompoundInterest {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter principal: ");
        double p = sc.nextDouble();
        System.out.print("Enter rate of interest (%): ");
        double r = sc.nextDouble();
        System.out.print("Enter number of years: ");
        int n = sc.nextInt();

        double amount = p * Math.pow(1 + r / 100, n);

        System.out.println("Amount after " + n + " years = " + String.format("%.2f", amount));
    }
}

Solved by 2 students in Fun Program Battle

6 Sum from 1 to N Easy for loop accumulator

Write a program that takes a positive integer n as input and calculates the sum 1 + 2 + 3 + ... + n. For example, if the input is 5 the output is 15. If the input is not positive, print an error message.

Sample input

Sample output

Sum = 15

Tip: The formula n * (n + 1) / 2 gives the same answer. Use it to check your loop.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class SumToN {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a positive integer: ");
        int n = sc.nextInt();

        if (n <= 0) {
            System.out.println("Please enter a positive integer.");
            return;
        }

        int sum = 0;
        for (int i = 1; i <= n; i++) {
            sum += i;               // add each number to the running total
        }
        System.out.println("Sum = " + sum);
    }
}

Solved by 5 students in Fun Program Battle

Topic 2 · 9 problems

Loops & Patterns

for and while loops, number series and patterns.

7 Fibonacci Series Easy while loop series

The Fibonacci series starts with 0 and 1, and every next term is the sum of the previous two (0, 1, 1, 2, 3, 5, 8, ...). Write a program that takes an integer n as input and prints all Fibonacci numbers that are less than or equal to n. For example, if n is 10 the output is 0 1 1 2 3 5 8.

Sample input

Sample output

0 1 1 2 3 5 8

Tip: You only need two variables for the last two terms. The third variable is just a temporary holder while shifting them forward.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class FibonacciUpToN {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter n: ");
        int n = sc.nextInt();

        int a = 0, b = 1;
        while (a <= n) {
            System.out.print(a + " ");
            int c = a + b;    // next term
            a = b;            // shift the window forward
            b = c;
        }
        System.out.println();
    }
}

Solved by 15 students in Fun Program Battle

8 Triangular Numbers Easy while loop series counting

A triangular number is the sum of the first k natural numbers: T(k) = k x (k + 1) / 2, giving 1, 3, 6, 10, 15, ... Write a program that takes an integer n and prints all triangular numbers up to n, followed by how many there are.

Sample input

Sample output

1 3 6 10 15
Count = 5

Tip: Each triangular number is the previous one plus the next k, so you can also build them as t = t + k without the formula.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class TriangularNumbers {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter n: ");
        int n = sc.nextInt();

        int k = 1, count = 0;
        int t = k * (k + 1) / 2;
        while (t <= n) {
            System.out.print(t + " ");
            count++;
            k++;
            t = k * (k + 1) / 2;    // next triangular number
        }
        System.out.println();
        System.out.println("Count = " + count);
    }
}

Solved by 12 students in Fun Program Battle

9 Perfect Squares Easy for loop series sum

A perfect square is the square of an integer: 1, 4, 9, 16, ... Write a program to read an integer n, print all perfect squares less than or equal to n separated by commas, and also print their sum.

Sample input

Sample output

1, 4, 9, 16, 25
Sum = 55

Tip: Loop over i (the root) instead of testing every number up to n. The condition i * i <= n stops the loop at the right time.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class PerfectSquares {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter n: ");
        int n = sc.nextInt();

        int sum = 0;
        for (int i = 1; i * i <= n; i++) {
            int sq = i * i;
            if (i > 1) {
                System.out.print(", ");    // comma before every square except the first
            }
            System.out.print(sq);
            sum += sq;
        }
        System.out.println();
        System.out.println("Sum = " + sum);
    }
}

Solved by 25 students in Fun Program Battle

10 Odd Squares Sum Easy for loop step of 2

Write a program that calculates the sum of the squares of all odd numbers less than a given integer N. For example, if N = 6 the odd numbers are 1, 3 and 5, so the output is 1 + 9 + 25 = 35.

Sample input

Sample output

Sum = 35

Tip: Start the loop at 1 and step by 2 (i += 2) so that it only visits odd numbers. There is no need for an if inside the loop.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class OddSquaresSum {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter N: ");
        int n = sc.nextInt();

        int sum = 0;
        for (int i = 1; i < n; i += 2) {   // visits only odd numbers
            sum += i * i;
        }
        System.out.println("Sum = " + sum);
    }
}

Solved by 2 students in Fun Program Battle

11 Factorial Digit Sum Medium factorial long digit extraction

Write a program that takes a non-negative integer n, calculates n! (n factorial), and then finds the sum of the digits of the result. For example, for n = 5 the factorial is 120 and the digit sum is 1 + 2 + 0 = 3. Accept values of n from 0 to 20.

Sample input

Sample output

Factorial = 120
Sum of digits = 3

Tip: Factorials grow very fast. int overflows after 12! and long after 20!, so this program uses long and limits n to 20. (For bigger values Java has BigInteger.) Remember that 0! = 1.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class FactorialDigitSum {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter n (0 to 20): ");
        int n = sc.nextInt();

        if (n < 0 || n > 20) {
            System.out.println("Please enter a number from 0 to 20.");
            return;
        }

        long fact = 1;
        for (int i = 1; i <= n; i++) {
            fact *= i;
        }

        long temp = fact;
        int sum = 0;
        while (temp > 0) {
            sum += (int) (temp % 10);    // last digit (always 0-9, so int is safe)
            temp /= 10;          // remove the last digit
        }

        System.out.println("Factorial = " + fact);
        System.out.println("Sum of digits = " + sum);
    }
}

Solved by 3 students in Fun Program Battle

12 Floyd's Triangle Easy nested loops patterns

Write a program that reads the number of rows n and prints a triangle of numbers where row 1 has one number, row 2 has two numbers, and so on. The numbers keep counting up from 1. For n = 4 the output is: 1 2 3 4 5 6 7 8 9 10

Sample input

Sample output

Tip: In almost every pattern program the outer loop controls the rows and the inner loop controls what is printed in each row.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class FloydTriangle {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter number of rows: ");
        int n = sc.nextInt();

        int num = 1;                          // the next number to print
        for (int i = 1; i <= n; i++) {        // one pass per row
            for (int j = 1; j <= i; j++) {    // row i has i numbers
                System.out.print(num + " ");
                num++;
            }
            System.out.println();
        }
    }
}

Solved by 6 students in Fun Program Battle

13 Growing Pattern Easy nested loops patterns

Write a program that reads a starting number s and the number of rows r, and prints a pattern where each row starts again from s and has one more number than the previous row. For s = 5 and r = 4 the output is: 5 5 6 5 6 7 5 6 7 8

Sample input

5
4

Sample output

Tip: Compare this with Floyd's triangle. There the counter keeps running across rows. Here it is reset to s at the start of every row.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class GrowingPattern {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter starting number: ");
        int s = sc.nextInt();
        System.out.print("Enter number of rows: ");
        int r = sc.nextInt();

        for (int i = 1; i <= r; i++) {
            for (int j = 0; j < i; j++) {
                System.out.print((s + j) + " ");    // every row starts again from s
            }
            System.out.println();
        }
    }
}

Solved by 6 students in Fun Program Battle

14 Odd Pyramid Medium nested loops patterns spacing

Write a program that takes the height h of a pyramid and prints consecutive odd numbers in a centred pyramid shape: row 1 has one odd number, row 2 has the next two, and so on. For h = 3 the output is: 1 3 5 7 9 11

Sample input

Sample output

  1
 3 5
7 9 11

Tip: Centring comes from printing (h - i) spaces before row i. Print the spaces first, then the numbers.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class OddPyramid {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter height: ");
        int h = sc.nextInt();

        int odd = 1;
        for (int i = 1; i <= h; i++) {
            for (int s = 1; s <= h - i; s++) {
                System.out.print(" ");           // leading spaces centre the row
            }
            for (int j = 1; j <= i; j++) {
                System.out.print(odd + " ");
                odd += 2;                         // next odd number
            }
            System.out.println();
        }
    }
}

Solved by 1 student in Fun Program Battle

15 Up and Down Easy for loop reverse loop

Write a program that takes an integer n and prints the numbers from 1 to n, followed by the same numbers in reverse order, all on one line. For example, if the input is 3 the output is 1 2 3 3 2 1.

Sample input

Sample output

1 2 3 3 2 1

Tip: A reverse loop starts at n, keeps going while i >= 1, and uses i--.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class UpAndDown {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter n: ");
        int n = sc.nextInt();

        for (int i = 1; i <= n; i++) {     // counting up
            System.out.print(i + " ");
        }
        for (int i = n; i >= 1; i--) {     // counting down
            System.out.print(i + " ");
        }
        System.out.println();
    }
}

Solved by 3 students in Fun Program Battle

Topic 3 · 24 problems

Digits

Take numbers apart with % 10 and / 10. The most popular battle topic.

16 Prime Digits Easy digit extraction while loop

Write a program that takes a positive integer and counts how many of its digits are prime. The prime digits are 2, 3, 5 and 7. For example, for 237 the output is 3, because all three digits are prime.

Sample input

Sample output

Prime digits = 3

Tip: This is the key digit-extraction trick used throughout this section: n % 10 gives the last digit and n / 10 removes it. Repeat until n becomes 0.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class PrimeDigitCount {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int n = Math.abs(sc.nextInt());

        int count = 0;
        while (n > 0) {
            int d = n % 10;                              // last digit
            if (d == 2 || d == 3 || d == 5 || d == 7) {
                count++;
            }
            n = n / 10;                                  // drop the last digit
        }
        System.out.println("Prime digits = " + count);
    }
}

Solved by 24 students in Fun Program Battle

17 Odd Digits Easy digit extraction modulus (%)

Write a program to count the number of odd digits in a given integer. For example, 12345 has three odd digits (1, 3 and 5), so the output is 3. For 24680 the output is 0.

Sample input

Sample output

Odd digits = 3

Tip: A digit d is odd when d % 2 != 0. Use Math.abs so that negative numbers work too.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class OddDigitCount {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int n = Math.abs(sc.nextInt());

        int count = 0;
        while (n > 0) {
            int d = n % 10;
            if (d % 2 != 0) {
                count++;
            }
            n /= 10;
        }
        System.out.println("Odd digits = " + count);
    }
}

Solved by 7 students in Fun Program Battle

18 Odd & Even Digits Easy digit extraction do-while

Write a program that takes a positive integer and counts how many of its digits are odd and how many are even. For example, for 1234 the output is: odd digits = 2, even digits = 2.

Sample input

Sample output

Odd digits = 2
Even digits = 2

Tip: A do-while loop runs at least once, so the input 0 is correctly counted as one even digit. A plain while (n > 0) loop would count nothing.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class OddEvenDigits {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int n = Math.abs(sc.nextInt());

        int odd = 0, even = 0;
        do {                       // do-while handles the input 0 correctly
            int d = n % 10;
            if (d % 2 == 0) {
                even++;
            } else {
                odd++;
            }
            n /= 10;
        } while (n > 0);

        System.out.println("Odd digits = " + odd);
        System.out.println("Even digits = " + even);
    }
}

Solved by 2 students in Fun Program Battle

19 Even Digit Sum Easy digit extraction sum

Write a program that calculates the sum of all even digits in a given integer. The number may be positive or negative. For example, for 123456 the output is 12 (2 + 4 + 6). If there are no even digits, the output is 0.

Sample input

Sample output

Sum of even digits = 12

Tip: 0 is an even digit too. It just adds nothing to the sum.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class EvenDigitSum {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int n = Math.abs(sc.nextInt());

        int sum = 0;
        while (n > 0) {
            int d = n % 10;
            if (d % 2 == 0) {
                sum += d;
            }
            n /= 10;
        }
        System.out.println("Sum of even digits = " + sum);
    }
}

Solved by 3 students in Fun Program Battle

20 Even Digit Product Easy digit extraction product flag

Write a program that takes a positive integer and calculates the product of all its even digits. If the number has no even digits, print 1. For example, for 123456 the output is 48 (2 x 4 x 6), and for 13579 the output is 1. A 0 digit is even, so it makes the product 0.

Sample input

Sample output

Product of even digits = 48

Tip: For a product, start at 1, not 0. Starting at 0 would make every answer 0.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class EvenDigitProduct {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int n = Math.abs(sc.nextInt());

        long product = 1;          // start a product at 1, not 0
        while (n > 0) {
            int d = n % 10;
            if (d % 2 == 0) {
                product *= d;
            }
            n /= 10;
        }
        System.out.println("Product of even digits = " + product);
    }
}

Solved by 3 students in Fun Program Battle

21 Digit Squares Sum Easy digit extraction sum

Write a program that takes a positive integer and calculates the sum of the squares of its digits. For example, for 123 the squares are 1, 4 and 9, so the output is 1 + 4 + 9 = 14.

Sample input

Sample output

Sum of squares of digits = 14

Tip: Multiply d * d rather than calling Math.pow(d, 2). It is simpler and stays an int.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class DigitSquareSum {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int n = Math.abs(sc.nextInt());

        int sum = 0;
        while (n > 0) {
            int d = n % 10;
            sum += d * d;
            n /= 10;
        }
        System.out.println("Sum of squares of digits = " + sum);
    }
}

Solved by 2 students in Fun Program Battle

22 Single-Digit Product Easy digit extraction product if-else

Write a program to find the product of the digits of a positive integer and check whether that product is a single-digit number. For example, for 123 the product is 1 x 2 x 3 = 6, which is a single digit. For 99 the product is 81, which is not.

Sample input

Sample output

Product of digits = 6
Single digit

Tip: A non-negative number is a single digit when it is less than 10.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class DigitProductCheck {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a positive integer: ");
        int n = sc.nextInt();

        int product = 1;
        int temp = n;
        do {
            product *= temp % 10;
            temp /= 10;
        } while (temp > 0);

        System.out.println("Product of digits = " + product);
        if (product < 10) {
            System.out.println("Single digit");
        } else {
            System.out.println("Not a single digit");
        }
    }
}

Solved by 8 students in Fun Program Battle

23 Digit Difference Easy digit extraction two sums

Write a program that splits the digits of an integer into two groups: digits that are multiples of 3 (0, 3, 6, 9) and all the others. Print (sum of the other digits) - (sum of the multiples of 3). For example, for 1234 the multiple of 3 is 3 and the others are 1 + 2 + 4 = 7, so the output is 7 - 3 = 4.

Sample input

Sample output

Difference = 4

Tip: Keep two separate running totals in one pass over the digits, then subtract at the end.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class DigitDifference {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int n = Math.abs(sc.nextInt());

        int mult3 = 0, others = 0;
        while (n > 0) {
            int d = n % 10;
            if (d % 3 == 0) {
                mult3 += d;
            } else {
                others += d;
            }
            n /= 10;
        }
        System.out.println("Difference = " + (others - mult3));
    }
}

Solved by 28 students in Fun Program Battle

24 Reverse & Sum Easy reverse number digit sum

Write a program that takes a positive integer and (1) reverses its digits and (2) finds the sum of its digits. Print both. If the input is not a positive integer, print an error message. For example, for 1234 the reversed number is 4321 and the digit sum is 10.

Sample input

Sample output

Reversed number = 4321
Sum of digits = 10

Tip: To reverse a number, rev = rev * 10 + digit shifts the digits collected so far one place left and adds the new digit at the end.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class ReverseAndSum {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a positive integer: ");
        int n = sc.nextInt();

        if (n <= 0) {
            System.out.println("Invalid input. Please enter a positive integer.");
            return;
        }

        int rev = 0, sum = 0;
        while (n > 0) {
            int d = n % 10;
            rev = rev * 10 + d;    // append d to the reversed number
            sum += d;
            n /= 10;
        }
        System.out.println("Reversed number = " + rev);
        System.out.println("Sum of digits = " + sum);
    }
}

Solved by 2 students in Fun Program Battle

25 Bigger Reverse Easy reverse number comparison

Write a program that takes a positive integer, reverses its digits, and checks whether the reversed number is greater than the original. For example, 123 reversed is 321, which is greater, so the program reports that the reverse is greater.

Sample input

Sample output

Reversed number = 321
The reversed number is greater

Tip: Keep a copy of the original number. The reversing loop destroys n as it goes.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class ReverseCompare {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a positive integer: ");
        int n = sc.nextInt();

        int orig = n;              // keep a copy before n is destroyed
        int rev = 0;
        while (n > 0) {
            rev = rev * 10 + n % 10;
            n /= 10;
        }

        System.out.println("Reversed number = " + rev);
        if (rev > orig) {
            System.out.println("The reversed number is greater");
        } else {
            System.out.println("The reversed number is not greater");
        }
    }
}

Solved by 3 students in Fun Program Battle

26 Palindrome Number Easy reverse number palindrome

A palindrome number reads the same forwards and backwards. Write a program that checks whether a given integer is a palindrome. For example, 121 is a palindrome, while 123 is not.

Sample input

Sample output

121 is a palindrome

Tip: A palindrome is just a number that equals its own reverse, so this reuses the reversing loop from the previous problems.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class PalindromeNumber {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int n = sc.nextInt();

        int temp = n, rev = 0;
        while (temp > 0) {
            rev = rev * 10 + temp % 10;
            temp /= 10;
        }

        if (rev == n) {
            System.out.println(n + " is a palindrome");
        } else {
            System.out.println(n + " is not a palindrome");
        }
    }
}

Solved by 9 students in Fun Program Battle

27 Digital Root Easy nested loops digit sum

The digital root of a number is found by adding its digits, then adding the digits of the result, and so on until a single digit remains. For example, the digital root of 16 is 1 + 6 = 7, and the digital root of 942 is 9 + 4 + 2 = 15, then 1 + 5 = 6. Write a program that prints the digital root of a given integer.

Sample input

Sample output

Digital root = 6

Tip: The outer loop repeats while the number has more than one digit (n > 9). The inner loop adds up the digits once.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class DigitalRoot {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int n = Math.abs(sc.nextInt());

        while (n > 9) {              // repeat until one digit is left
            int sum = 0;
            while (n > 0) {          // add the digits once
                sum += n % 10;
                n /= 10;
            }
            n = sum;
        }
        System.out.println("Digital root = " + n);
    }
}

Solved by 22 students in Fun Program Battle

28 Magic Number Easy digital root nested loops

A magic number is a number that becomes 1 when you keep adding its digits until a single digit is left. For example, 19 is a magic number because 1 + 9 = 10 and 1 + 0 = 1. Write a program that checks whether a given number is a magic number and shows each step of the reduction.

Sample input

Sample output

19 -> 10 -> 1
19 is a Magic Number

Tip: This is the definition used in ICSE. A magic number is simply a number whose digital root is 1.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class MagicNumber {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int orig = sc.nextInt();

        int n = orig;
        System.out.print(n);
        while (n > 9) {
            int sum = 0;
            while (n > 0) {
                sum += n % 10;
                n /= 10;
            }
            n = sum;
            System.out.print(" -> " + n);    // show each step
        }
        System.out.println();

        if (n == 1) {
            System.out.println(orig + " is a Magic Number");
        } else {
            System.out.println(orig + " is not a Magic Number");
        }
    }
}

Solved by 20 students in Fun Program Battle

29 Digit Product Chain Medium nested loops digit product

Write a program that takes a positive integer and multiplies its digits. If the product has more than one digit, multiply the digits of the product, and repeat until a single digit is left. Print each step and the final digit. For example, 39 gives 3 x 9 = 27, then 2 x 7 = 14, then 1 x 4 = 4, so the output is 4.

Sample input

Sample output

39 -> 27 -> 14 -> 4
Result = 4

Tip: This is the same structure as the digital root, but with multiplication. Remember to start each product at 1.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class DigitProductChain {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a positive integer: ");
        int n = sc.nextInt();

        System.out.print(n);
        while (n > 9) {
            int product = 1;
            while (n > 0) {
                product *= n % 10;
                n /= 10;
            }
            n = product;
            System.out.print(" -> " + n);
        }
        System.out.println();
        System.out.println("Result = " + n);
    }
}

Solved by 2 students in Fun Program Battle

30 Digit Frequency Medium arrays as counters digit extraction

Write a program that takes a positive integer n and prints how many times each digit 0 to 9 occurs in it, one digit per line. If n is less than 1, print an error message. For example, for 122333 the output shows 1: 1, 2: 2, 3: 3, and 0 for every other digit.

Sample input

Sample output

0: 0
1: 1
2: 2
3: 3
4: 0
5: 0
6: 0
7: 0
8: 0
9: 0

Tip: An array int[10] works as ten counters at once. count[d]++ increases the counter for digit d.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class DigitFrequency {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a positive integer: ");
        int n = sc.nextInt();

        if (n < 1) {
            System.out.println("Please enter a positive integer.");
            return;
        }

        int[] count = new int[10];       // count[d] = how often digit d appears
        while (n > 0) {
            count[n % 10]++;
            n /= 10;
        }
        for (int d = 0; d <= 9; d++) {
            System.out.println(d + ": " + count[d]);
        }
    }
}

Solved by 2 students in Fun Program Battle

31 Triple Digits Medium digit extraction tracking previous value

Write a program that checks whether a positive integer contains three or more identical digits in a row. For example, 122333 contains "333", so the output is "Yes". 112233 has no such run, so the output is "No".

Sample input

Sample output

Yes

Tip: Remember the previous digit and the length of the current run. When the new digit equals the previous one the run grows, otherwise it restarts at 1.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class TripleDigits {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a positive integer: ");
        int n = sc.nextInt();

        int prev = -1, run = 0;
        boolean found = false;
        while (n > 0) {
            int d = n % 10;
            if (d == prev) {
                run++;              // same digit again: the run grows
            } else {
                run = 1;            // a new digit starts a new run
            }
            if (run >= 3) {
                found = true;
            }
            prev = d;
            n /= 10;
        }
        System.out.println(found ? "Yes" : "No");
    }
}

Solved by 6 students in Fun Program Battle

32 Target Digit Sum Medium search loop break digit sum

Write a program to find the smallest positive number whose digits add up to N. Search the numbers from 1 to 1,000,000. For example, for N = 10 the answer is 19 (1 + 9 = 10). If no such number exists in the range, print a message.

Sample input

Sample output

Smallest number = 19

Tip: Search from 1 upwards and break at the first match, which is automatically the smallest. The largest digit sum below 1,000,000 is 54 (999999), so a larger N has no answer.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class SmallestWithDigitSum {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter N: ");
        int target = sc.nextInt();

        int answer = -1;
        for (int i = 1; i <= 1000000; i++) {
            int t = i, sum = 0;
            while (t > 0) {
                sum += t % 10;
                t /= 10;
            }
            if (sum == target) {
                answer = i;
                break;              // the first match is the smallest
            }
        }

        if (answer == -1) {
            System.out.println("No such number up to 1000000");
        } else {
            System.out.println("Smallest number = " + answer);
        }
    }
}

Solved by 46 students in Fun Program Battle

33 Multiples Digit Sum Easy for loop if digit sum

Write a program that takes three integers A, B and K, and finds the total sum of the digits of all numbers from A to B (inclusive) that are divisible by K. For example, for A = 1, B = 10 and K = 2 the numbers are 2, 4, 6, 8 and 10, so the total is 2 + 4 + 6 + 8 + (1 + 0) = 21.

Sample input

1
10
2

Sample output

Total digit sum = 21

Tip: Copy the loop variable into a temporary variable before extracting its digits. Changing i itself would break the loop.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class MultipleDigitSum {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter A: ");
        int a = sc.nextInt();
        System.out.print("Enter B: ");
        int b = sc.nextInt();
        System.out.print("Enter K: ");
        int k = sc.nextInt();

        int total = 0;
        for (int i = a; i <= b; i++) {
            if (i % k == 0) {
                int t = Math.abs(i);    // work on a copy, never on i itself
                while (t > 0) {
                    total += t % 10;
                    t /= 10;
                }
            }
        }
        System.out.println("Total digit sum = " + total);
    }
}

Solved by 63 students in Fun Program Battle

34 Lucky Numbers Easy for loop digit sum multiples

In this problem a number is called "lucky" if the sum of its digits is a multiple of 8. For example, 17 is lucky because 1 + 7 = 8. Write a program that takes a start and an end value and prints all lucky numbers between them (inclusive).

Sample input

1
50

Sample output

8 17 26 35 44

Tip: "Lucky" is a made-up rule for this problem only. ICSE also has a different, sieve-based "lucky number", so don't mix the two up.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class LuckyNumbers {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter start: ");
        int start = sc.nextInt();
        System.out.print("Enter end: ");
        int end = sc.nextInt();

        for (int i = Math.max(start, 1); i <= end; i++) {
            int t = i, sum = 0;
            while (t > 0) {
                sum += t % 10;
                t /= 10;
            }
            if (sum % 8 == 0) {
                System.out.print(i + " ");
            }
        }
        System.out.println();
    }
}

Solved by 29 students in Fun Program Battle

35 Armstrong Number Medium counting digits powers

An Armstrong number is a number that equals the sum of its digits, each raised to the power of the number of digits. For example, 153 has 3 digits and 1^3 + 5^3 + 3^3 = 153, so it is an Armstrong number. Write a program that checks whether a given positive integer is an Armstrong number.

Sample input

Sample output

153 is an Armstrong number

Tip: You need two passes over the digits: first to count them, then to add the powers. Raising to a power with a small loop avoids the double rounding that Math.pow can bring.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class Armstrong {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a positive integer: ");
        int n = sc.nextInt();

        // Pass 1: count the digits
        int k = 0;
        for (int t = n; t > 0; t /= 10) {
            k++;
        }

        // Pass 2: add each digit raised to the power k
        int sum = 0;
        for (int t = n; t > 0; t /= 10) {
            int d = t % 10, power = 1;
            for (int i = 1; i <= k; i++) {
                power *= d;
            }
            sum += power;
        }

        if (sum == n) {
            System.out.println(n + " is an Armstrong number");
        } else {
            System.out.println(n + " is not an Armstrong number");
        }
    }
}

Solved by 2 students in Fun Program Battle

36 Special Number Medium factorial String digits

A Special number (also called a Krishnamurthy number) equals the sum of the factorials of its digits. For example, 145 = 1! + 4! + 5! = 1 + 24 + 120. Write a program that takes a positive integer, shows the factorial sum of its digits, and says whether it is a Special number.

Sample input

Sample output

1! + 4! + 5! = 145
Yes, 145 is a Special number

Tip: Converting the number to a String lets you read its digits from left to right with charAt(i) - '0'. This keeps the printed breakdown in the right order.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class SpecialNumber {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a positive integer: ");
        int n = sc.nextInt();

        String s = String.valueOf(n);
        int sum = 0;
        for (int i = 0; i < s.length(); i++) {
            int d = s.charAt(i) - '0';      // character '4' becomes digit 4
            int fact = 1;
            for (int j = 2; j <= d; j++) {
                fact *= j;
            }
            sum += fact;
            System.out.print((i > 0 ? " + " : "") + d + "!");
        }
        System.out.println(" = " + sum);

        if (sum == n) {
            System.out.println("Yes, " + n + " is a Special number");
        } else {
            System.out.println("No, " + n + " is not a Special number");
        }
    }
}

Solved by 1 student in Fun Program Battle

37 Swap & Prime Easy two-digit numbers prime check

Write a program that reads a two-digit number, swaps its two digits, and checks whether the new number is prime. For example, 13 becomes 31, which is prime. 12 becomes 21 = 3 x 7, which is not prime.

Sample input

Sample output

Swapped number = 31
31 is prime

Tip: For a two-digit number, n / 10 is the tens digit and n % 10 is the units digit. Note that 1 is not a prime number.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class SwapAndPrime {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a two-digit number: ");
        int n = sc.nextInt();

        if (n < 10 || n > 99) {
            System.out.println("Please enter a two-digit number.");
            return;
        }

        int swapped = (n % 10) * 10 + n / 10;
        boolean prime = swapped > 1;
        for (int i = 2; i * i <= swapped; i++) {
            if (swapped % i == 0) {
                prime = false;
                break;
            }
        }

        System.out.println("Swapped number = " + swapped);
        System.out.println(swapped + (prime ? " is prime" : " is not prime"));
    }
}

Solved by 3 students in Fun Program Battle

38 Next Palindromes Medium palindrome counting loop

Write a program that takes a starting number and a count N, and prints the first N palindromic numbers that are greater than or equal to the starting number. A palindromic number reads the same when reversed. For example, from 123 with N = 5 the output is 131, 141, 151, 161 and 171.

Sample input

123
5

Sample output

131 141 151 161 171

Tip: When you don't know in advance how many numbers you will test, use a while loop that stops once you have found N matches.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class NextPalindromes {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter starting number: ");
        int k = sc.nextInt();
        System.out.print("Enter N: ");
        int n = sc.nextInt();

        int found = 0;
        while (found < n) {
            int rev = 0;
            for (int t = k; t > 0; t /= 10) {
                rev = rev * 10 + t % 10;
            }
            if (rev == k) {
                System.out.print(k + " ");
                found++;
            }
            k++;
        }
        System.out.println();
    }
}

Solved by 4 students in Fun Program Battle

39 Unique Digit Numbers Medium boolean array nested loops

Write a program to count the numbers from 1 to N whose digits are all different (no digit repeats). For example, for N = 25 every number qualifies except 11 and 22, so the output is 23.

Sample input

Sample output

Count = 23

Tip: A boolean array seen[10] records which digits have already appeared. Make a fresh array for every number you test.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class UniqueDigitNumbers {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter N: ");
        int n = sc.nextInt();

        int count = 0;
        for (int m = 1; m <= n; m++) {
            boolean[] seen = new boolean[10];    // fresh for each number
            boolean repeat = false;
            for (int t = m; t > 0; t /= 10) {
                int d = t % 10;
                if (seen[d]) {
                    repeat = true;
                    break;
                }
                seen[d] = true;
            }
            if (!repeat) {
                count++;
            }
        }
        System.out.println("Count = " + count);
    }
}

Solved by 10 students in Fun Program Battle

Topic 4 · 8 problems

Primes & Divisors

Prime checks, prime factors and divisors.

40 Primes in a Range Easy prime check nested loops

Write a program that takes two integers, start and end, and counts the prime numbers in the range [start, end]. For example, between 10 and 20 there are 4 primes (11, 13, 17 and 19), so the output is 4.

Sample input

10
20

Sample output

Count of primes = 4

Tip: To test whether n is prime you only need to try divisors up to the square root of n (i * i <= n). Any larger factor would pair with a smaller one you have already tried.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class PrimeCount {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter start: ");
        int start = sc.nextInt();
        System.out.print("Enter end: ");
        int end = sc.nextInt();

        int count = 0;
        for (int n = start; n <= end; n++) {
            boolean prime = n > 1;               // 0, 1 and negatives are not prime
            for (int i = 2; i * i <= n; i++) {
                if (n % i == 0) {
                    prime = false;
                    break;
                }
            }
            if (prime) {
                count++;
            }
        }
        System.out.println("Count of primes = " + count);
    }
}

Solved by 54 students in Fun Program Battle

41 Prime Product Medium prime check long product

Write a program to read an integer n, print all prime numbers less than or equal to n in ascending order separated by commas, and also print the product of those primes. For example, for n = 10 the primes are 2, 3, 5, 7 and their product is 2 x 3 x 5 x 7 = 210.

Sample input

Sample output

2, 3, 5, 7
Product = 210

Tip: Products grow quickly, so store the product in a long.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class PrimeProduct {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter n: ");
        int n = sc.nextInt();

        long product = 1;
        boolean first = true;
        for (int k = 2; k <= n; k++) {
            boolean prime = true;
            for (int i = 2; i * i <= k; i++) {
                if (k % i == 0) {
                    prime = false;
                    break;
                }
            }
            if (prime) {
                System.out.print((first ? "" : ", ") + k);
                first = false;
                product *= k;
            }
        }
        System.out.println();
        System.out.println("Product = " + product);
    }
}

Solved by 3 students in Fun Program Battle

42 Prime Sum Easy prime check sum

Write a program to find the sum of all prime numbers less than a given number N. For example, for N = 10 the primes below 10 are 2, 3, 5 and 7, so the output is 2 + 3 + 5 + 7 = 17.

Sample input

Sample output

Sum of primes = 17

Tip: Read the question carefully: "less than N" means the loop condition is k < N, not k <= N.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class PrimeSum {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter N: ");
        int n = sc.nextInt();

        int sum = 0;
        for (int k = 2; k < n; k++) {
            boolean prime = true;
            for (int i = 2; i * i <= k; i++) {
                if (k % i == 0) {
                    prime = false;
                    break;
                }
            }
            if (prime) {
                sum += k;
            }
        }
        System.out.println("Sum of primes = " + sum);
    }
}

Solved by 2 students in Fun Program Battle

43 Prime Digit Sum Easy digit sum prime check

Write a program that checks whether the sum of the digits of a given number is a prime number. For example, for 23 the digit sum is 2 + 3 = 5, which is prime. For 123 the digit sum is 6 and for 37 it is 10, and neither is prime.

Sample input

Sample output

Sum of digits = 5
The digit sum is prime

Tip: Break the problem into two small jobs you already know, a digit sum and a prime check, and do them one after the other.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class DigitSumPrime {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int n = Math.abs(sc.nextInt());

        int sum = 0;
        while (n > 0) {
            sum += n % 10;
            n /= 10;
        }

        boolean prime = sum > 1;
        for (int i = 2; i * i <= sum; i++) {
            if (sum % i == 0) {
                prime = false;
                break;
            }
        }

        System.out.println("Sum of digits = " + sum);
        System.out.println(prime ? "The digit sum is prime" : "The digit sum is not prime");
    }
}

Solved by 4 students in Fun Program Battle

44 Perfect Number Easy divisors sum

A perfect number is a positive integer equal to the sum of its proper divisors (all divisors except the number itself). For example, the proper divisors of 6 are 1, 2 and 3, and 1 + 2 + 3 = 6, so 6 is perfect. Write a program that checks whether a given number is perfect.

Sample input

Sample output

6 is a perfect number

Tip: No proper divisor can be bigger than n / 2, so the loop can stop there. 1 is not a perfect number, because it has no proper divisors.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class PerfectNumber {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a positive integer: ");
        int n = sc.nextInt();

        int sum = 0;
        for (int i = 1; i <= n / 2; i++) {    // proper divisors are at most n / 2
            if (n % i == 0) {
                sum += i;
            }
        }

        if (n > 1 && sum == n) {
            System.out.println(n + " is a perfect number");
        } else {
            System.out.println(n + " is not a perfect number");
        }
    }
}

Solved by 8 students in Fun Program Battle

45 Prime Factor Sum Medium prime factorisation while loop

Write a program to find the sum of the distinct prime factors of a given number N. For example, 10 = 2 x 5, so the output is 2 + 5 = 7. 12 = 2 x 2 x 3, so the distinct prime factors are 2 and 3 and the output is 5. If N is less than 2, print a message.

Sample input

Sample output

Prime factors: 2 5
Sum = 7

Tip: Dividing out each factor completely (while n % i == 0, n = n / i) guarantees that every i which divides n is a prime. Whatever is left above 1 at the end is also a prime factor.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class PrimeFactorSum {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter N: ");
        int n = sc.nextInt();

        if (n < 2) {
            System.out.println("N must be 2 or more.");
            return;
        }

        int sum = 0;
        System.out.print("Prime factors:");
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) {
                System.out.print(" " + i);
                sum += i;
                while (n % i == 0) {
                    n /= i;                  // remove this factor completely
                }
            }
        }
        if (n > 1) {                         // a prime factor bigger than sqrt(N) is left
            System.out.print(" " + n);
            sum += n;
        }
        System.out.println();
        System.out.println("Sum = " + sum);
    }
}

Solved by 10 students in Fun Program Battle

46 K Divisors Medium counting divisors nested loops

Write a program that takes two integers N and K and counts the numbers less than N that have exactly K divisors. For example, for N = 10 and K = 2 the numbers 2, 3, 5 and 7 each have exactly 2 divisors, so the output is 4.

Sample input

10
2

Sample output

Numbers: 2 3 5 7
Count = 4

Tip: Numbers with exactly 2 divisors are the primes. Numbers with exactly 3 divisors are squares of primes (4, 9, 25, ...). Try different values of K and look for the pattern.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class KDivisors {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter N: ");
        int n = sc.nextInt();
        System.out.print("Enter K: ");
        int k = sc.nextInt();

        int answer = 0;
        System.out.print("Numbers:");
        for (int m = 1; m < n; m++) {
            int divisors = 0;
            for (int i = 1; i <= m; i++) {
                if (m % i == 0) {
                    divisors++;
                }
            }
            if (divisors == k) {
                System.out.print(" " + m);
                answer++;
            }
        }
        System.out.println();
        System.out.println("Count = " + answer);
    }
}

Solved by 6 students in Fun Program Battle

47 Divisor Digit Sum Medium divisors digit sum

Write a program that takes an integer, finds all of its divisors, and adds up the digit sums of those divisors. For example, the divisors of 12 are 1, 2, 3, 4, 6 and 12. Their digit sums are 1, 2, 3, 4, 6 and 1 + 2 = 3, so the total is 19.

Sample input

Sample output

Divisors: 1 2 3 4 6 12
Total = 19

Tip: Each divisor must be copied into a temporary variable before you extract its digits. Changing the loop variable itself would break the loop.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class DivisorDigitSum {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int n = sc.nextInt();

        int total = 0;
        System.out.print("Divisors:");
        for (int i = 1; i <= n; i++) {
            if (n % i == 0) {
                System.out.print(" " + i);
                for (int t = i; t > 0; t /= 10) {    // digits of the divisor
                    total += t % 10;
                }
            }
        }
        System.out.println();
        System.out.println("Total = " + total);
    }
}

Solved by 4 students in Fun Program Battle

Topic 5 · 14 problems

Arrays

Search, count, swap, sort and rotate a list of values.

48 Swap Pairs Easy arrays swapping

Write a program to read an array of integers and swap each pair of neighbouring elements: the element at index 0 with index 1, index 2 with index 3, and so on. If the array has an odd length, the last element stays where it is. For example, [1, 2, 3, 4, 5] becomes [2, 1, 4, 3, 5].

Sample input

5
1 2 3 4 5

Sample output

2 1 4 3 5

Tip: Swapping always needs a third variable: temp = a[i]; a[i] = a[i+1]; a[i+1] = temp. Stepping i by 2 visits each pair once.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class SwapPairs {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter number of elements: ");
        int n = sc.nextInt();
        int[] a = new int[n];
        System.out.print("Enter the elements: ");
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }

        for (int i = 0; i + 1 < n; i += 2) {
            int temp = a[i];          // classic three-step swap
            a[i] = a[i + 1];
            a[i + 1] = temp;
        }

        for (int i = 0; i < n; i++) {
            System.out.print(a[i] + " ");
        }
        System.out.println();
    }
}

Solved by 11 students in Fun Program Battle

49 Alternate Sum Easy arrays index parity

Write a program that reads a list of integers and calculates (sum of elements at even indices) - (sum of elements at odd indices). Indices start at 0. For example, for [1, 2, 3, 4, 5] the output is (1 + 3 + 5) - (2 + 4) = 3. An empty list gives 0.

Sample input

5
1 2 3 4 5

Sample output

Result = 3

Tip: The index decides the group, not the value. Check i % 2, not a[i] % 2.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class AlternateSum {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter number of elements: ");
        int n = sc.nextInt();
        int[] a = new int[n];
        System.out.print("Enter the elements: ");
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }

        int result = 0;
        for (int i = 0; i < n; i++) {
            if (i % 2 == 0) {
                result += a[i];      // even index
            } else {
                result -= a[i];      // odd index
            }
        }
        System.out.println("Result = " + result);
    }
}

Solved by 2 students in Fun Program Battle

50 Peak Element Easy arrays neighbours boundary checks

A peak element is an element that is not smaller than its neighbours. Write a program to find the index of the first peak element in an array. The array has at least one element. For example, in [1, 3, 20, 4, 1, 0] the output is 2, because 20 is the peak.

Sample input

6
1 3 20 4 1 0

Sample output

Peak element 20 at index 2

Tip: The first and last elements have only one neighbour each. Check i > 0 before looking at a[i - 1], and i < n - 1 before looking at a[i + 1].

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class PeakElement {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter number of elements: ");
        int n = sc.nextInt();
        int[] a = new int[n];
        System.out.print("Enter the elements: ");
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }

        for (int i = 0; i < n; i++) {
            boolean leftOk = (i == 0) || a[i] >= a[i - 1];
            boolean rightOk = (i == n - 1) || a[i] >= a[i + 1];
            if (leftOk && rightOk) {
                System.out.println("Peak element " + a[i] + " at index " + i);
                break;
            }
        }
    }
}

Solved by 8 students in Fun Program Battle

51 Pair Sum Easy arrays nested loops

Write a program to read an array of integers and a target sum, then print every pair of values that adds up to the target. For example, for the array [1, 2, 3, 4, 5] and target 7 the output is (2, 5) and (3, 4). If there is no such pair, print a message.

Sample input

5
1 2 3 4 5
7

Sample output

(2, 5)
(3, 4)

Tip: Starting the inner loop at j = i + 1 checks each pair only once and never pairs an element with itself.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class PairSum {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter number of elements: ");
        int n = sc.nextInt();
        int[] a = new int[n];
        System.out.print("Enter the elements: ");
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        System.out.print("Enter target sum: ");
        int target = sc.nextInt();

        boolean found = false;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {     // j > i: each pair once
                if (a[i] + a[j] == target) {
                    System.out.println("(" + a[i] + ", " + a[j] + ")");
                    found = true;
                }
            }
        }
        if (!found) {
            System.out.println("No pairs found");
        }
    }
}

Solved by 4 students in Fun Program Battle

52 Element Frequency Medium arrays bubble sort counting

Write a program to read N integers into an array and print how many times each different value occurs, in ascending order of value. The input does not have to be sorted. For example, for [4, 2, 2, 3, 4, 4, 1] the output is 1: 1, 2: 2, 3: 1, 4: 3.

Sample input

7
4 2 2 3 4 4 1

Sample output

1: 1
2: 2
3: 1
4: 3

Tip: After sorting, equal values sit next to each other, so you can count each group in one pass. This program uses bubble sort, the sorting method taught in ICSE.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class ElementFrequency {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter number of elements: ");
        int n = sc.nextInt();
        int[] a = new int[n];
        System.out.print("Enter the elements: ");
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }

        // Bubble sort: equal values end up next to each other
        for (int pass = 0; pass < n - 1; pass++) {
            for (int i = 0; i < n - 1 - pass; i++) {
                if (a[i] > a[i + 1]) {
                    int temp = a[i];
                    a[i] = a[i + 1];
                    a[i + 1] = temp;
                }
            }
        }

        int count = 1;
        for (int i = 1; i <= n; i++) {
            if (i < n && a[i] == a[i - 1]) {
                count++;
            } else {
                System.out.println(a[i - 1] + ": " + count);    // run of a[i-1] has ended
                count = 1;
            }
        }
    }
}

Solved by 3 students in Fun Program Battle

53 Remove Duplicates Medium arrays search building a new array

Write a program that reads an array of integers and builds a new array containing only the first occurrence of each value, keeping the original order. Do not use any built-in function for removing duplicates. For example, [1, 2, 2, 3, 4, 1, 5] becomes [1, 2, 3, 4, 5].

Sample input

7
1 2 2 3 4 1 5

Sample output

1 2 3 4 5

Tip: Keep a counter m for how many values are in the result so far. The result array can be the same size as the input, because you only use its first m slots.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class RemoveDuplicates {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter number of elements: ");
        int n = sc.nextInt();
        int[] a = new int[n];
        System.out.print("Enter the elements: ");
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }

        int[] result = new int[n];
        int m = 0;                              // how many unique values so far
        for (int i = 0; i < n; i++) {
            boolean seen = false;
            for (int j = 0; j < m; j++) {
                if (result[j] == a[i]) {
                    seen = true;
                    break;
                }
            }
            if (!seen) {
                result[m] = a[i];
                m++;
            }
        }

        for (int i = 0; i < m; i++) {
            System.out.print(result[i] + " ");
        }
        System.out.println();
    }
}

Solved by 2 students in Fun Program Battle

54 Evens First Easy arrays two passes

Write a program that reads an array and rearranges it so that all even numbers come before all odd numbers, keeping the original order within each group. For example, [3, 1, 2, 4, 5] becomes [2, 4, 3, 1, 5].

Sample input

5
3 1 2 4 5

Sample output

2 4 3 1 5

Tip: Two simple passes over the array keep the order stable. The first pass copies the evens and the second copies the odds.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class EvensFirst {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter number of elements: ");
        int n = sc.nextInt();
        int[] a = new int[n];
        System.out.print("Enter the elements: ");
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }

        int[] b = new int[n];
        int k = 0;
        for (int i = 0; i < n; i++) {      // pass 1: even numbers
            if (a[i] % 2 == 0) {
                b[k++] = a[i];
            }
        }
        for (int i = 0; i < n; i++) {      // pass 2: odd numbers
            if (a[i] % 2 != 0) {
                b[k++] = a[i];
            }
        }

        for (int i = 0; i < n; i++) {
            System.out.print(b[i] + " ");
        }
        System.out.println();
    }
}

Solved by 1 student in Fun Program Battle

55 Longest Run Easy arrays counting runs maximum

Write a program that reads an array of integers and finds the length of the longest run of identical elements next to each other. For example, in [1, 2, 2, 3, 3, 3, 4, 4, 4, 4] the number 4 appears four times in a row, so the output is 4.

Sample input

10
1 2 2 3 3 3 4 4 4 4

Sample output

Longest run = 4

Tip: This combines two common patterns: counting the current run and keeping track of the maximum so far.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class LongestRun {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter number of elements: ");
        int n = sc.nextInt();
        int[] a = new int[n];
        System.out.print("Enter the elements: ");
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }

        int run = 1, best = 1;
        for (int i = 1; i < n; i++) {
            if (a[i] == a[i - 1]) {
                run++;
            } else {
                run = 1;
            }
            if (run > best) {
                best = run;
            }
        }
        System.out.println("Longest run = " + best);
    }
}

Solved by 1 student in Fun Program Battle

56 Reverse Array Easy arrays swapping palindrome

Write a program that reads an array of integers and reverses it in place (without a second array for the result). Then check whether the reversed array is the same as the original. If it is, print "Palindrome Array", otherwise print the reversed array.

Sample input

4
1 2 3 4

Sample output

Reversed array: 4 3 2 1

Tip: To reverse in place, swap the first and last elements, then the second and second-last, and so on. Stop at the middle (i < n / 2), or you will swap everything back.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class ReverseArray {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter number of elements: ");
        int n = sc.nextInt();
        int[] a = new int[n];
        int[] original = new int[n];
        System.out.print("Enter the elements: ");
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
            original[i] = a[i];
        }

        for (int i = 0; i < n / 2; i++) {        // swap ends, moving inwards
            int temp = a[i];
            a[i] = a[n - 1 - i];
            a[n - 1 - i] = temp;
        }

        boolean same = true;
        for (int i = 0; i < n; i++) {
            if (a[i] != original[i]) {
                same = false;
                break;
            }
        }

        if (same) {
            System.out.println("Palindrome Array");
        } else {
            System.out.print("Reversed array: ");
            for (int i = 0; i < n; i++) {
                System.out.print(a[i] + " ");
            }
            System.out.println();
        }
    }
}

Solved by 1 student in Fun Program Battle

57 Pivot Index Medium arrays running sum

Write a program to find the pivot index of an array: the index where the sum of the elements to its left equals the sum of the elements to its right. Print -1 if no such index exists. For example, in [1, 2, 3, 2, 1] the output is 2, because 1 + 2 = 2 + 1.

Sample input

5
1 2 3 2 1

Sample output

Pivot index = 2

Tip: Instead of re-adding both sides for every index, compute the total once. Then right = total - left - a[i], which is much faster.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class PivotIndex {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter number of elements: ");
        int n = sc.nextInt();
        int[] a = new int[n];
        System.out.print("Enter the elements: ");
        int total = 0;
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
            total += a[i];
        }

        int left = 0, pivot = -1;
        for (int i = 0; i < n; i++) {
            int right = total - left - a[i];
            if (left == right) {
                pivot = i;
                break;
            }
            left += a[i];
        }
        System.out.println("Pivot index = " + pivot);
    }
}

Solved by 1 student in Fun Program Battle

58 Peaks & Valleys Easy arrays neighbours

Write a program that reads an array of integers and counts its peaks and valleys. A peak is an element greater than both neighbours, and a valley is an element smaller than both neighbours. The first and last elements are never counted. For example, in [1, 3, 2, 4, 1] the peaks are 3 and 4 and the valley is 2, so the total is 3. An array with fewer than 3 elements has none.

Sample input

5
1 3 2 4 1

Sample output

Peaks = 2
Valleys = 1
Total = 3

Tip: Loop only from i = 1 to n - 2, so that both a[i - 1] and a[i + 1] always exist.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class PeaksAndValleys {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter number of elements: ");
        int n = sc.nextInt();
        int[] a = new int[n];
        System.out.print("Enter the elements: ");
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }

        int peaks = 0, valleys = 0;
        for (int i = 1; i < n - 1; i++) {        // skip the two ends
            if (a[i] > a[i - 1] && a[i] > a[i + 1]) {
                peaks++;
            } else if (a[i] < a[i - 1] && a[i] < a[i + 1]) {
                valleys++;
            }
        }
        System.out.println("Peaks = " + peaks);
        System.out.println("Valleys = " + valleys);
        System.out.println("Total = " + (peaks + valleys));
    }
}

Solved by 1 student in Fun Program Battle

59 Array Median Medium arrays bubble sort odd/even length

Write a program that reads an array of integers and prints its median: the middle value after sorting. If the array has an even number of elements, the median is the average of the two middle values. For example, the median of [1, 3, 2, 4, 5] is 3, and the median of [1, 2, 3, 4] is 2.5. Print a message if the array is empty.

Sample input

5
1 3 2 4 5

Sample output

Median = 3.0

Tip: Divide by 2.0, not 2, when averaging. Integer division would turn 2.5 into 2.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class Median {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter number of elements: ");
        int n = sc.nextInt();
        if (n == 0) {
            System.out.println("Array is empty");
            return;
        }
        int[] a = new int[n];
        System.out.print("Enter the elements: ");
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }

        for (int pass = 0; pass < n - 1; pass++) {      // bubble sort
            for (int i = 0; i < n - 1 - pass; i++) {
                if (a[i] > a[i + 1]) {
                    int temp = a[i];
                    a[i] = a[i + 1];
                    a[i + 1] = temp;
                }
            }
        }

        double median;
        if (n % 2 == 1) {
            median = a[n / 2];
        } else {
            median = (a[n / 2 - 1] + a[n / 2]) / 2.0;      // 2.0 keeps the decimal part
        }
        System.out.println("Median = " + median);
    }
}

Solved by 5 students in Fun Program Battle

60 Rotation Check Medium arrays modulus (%) rotation

Write a program that reads two arrays of the same length and checks whether the second is a rotation of the first. For example, [3, 4, 5, 1, 2] is a rotation of [1, 2, 3, 4, 5] (shifted by 2 places), but [1, 3, 2] is not a rotation of [1, 2, 3].

Sample input

5
1 2 3 4 5
3 4 5 1 2

Sample output

The second array is a rotation of the first

Tip: The index (i + shift) % n wraps around to the start of the array. This is how you treat an array as a circle.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class RotationCheck {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter number of elements: ");
        int n = sc.nextInt();
        int[] a = new int[n];
        int[] b = new int[n];
        System.out.print("Enter the first array: ");
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        System.out.print("Enter the second array: ");
        for (int i = 0; i < n; i++) {
            b[i] = sc.nextInt();
        }

        boolean rotation = false;
        for (int shift = 0; shift < n && !rotation; shift++) {
            boolean match = true;
            for (int i = 0; i < n; i++) {
                if (b[i] != a[(i + shift) % n]) {   // % n wraps around the end
                    match = false;
                    break;
                }
            }
            rotation = match;
        }

        if (rotation) {
            System.out.println("The second array is a rotation of the first");
        } else {
            System.out.println("The second array is not a rotation of the first");
        }
    }
}

Solved by 9 students in Fun Program Battle

61 Missing Number Easy arrays consecutive values

Write a program that reads a sorted list of consecutive integers with exactly one number missing from the middle, and prints the missing number. For example, for [1, 2, 3, 5, 6] the output is 4.

Sample input

5
1 2 3 5 6

Sample output

Missing number = 4

Tip: In a consecutive list each element is exactly one more than the one before it. The first place this fails shows you the gap.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class MissingNumber {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter number of elements: ");
        int n = sc.nextInt();
        int[] a = new int[n];
        System.out.print("Enter the elements: ");
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }

        for (int i = 1; i < n; i++) {
            if (a[i] != a[i - 1] + 1) {          // the gap is right here
                System.out.println("Missing number = " + (a[i - 1] + 1));
                break;
            }
        }
    }
}

Solved by 6 students in Fun Program Battle

Topic 6 · 10 problems

Strings

Vowels, palindromes, anagrams and word tricks.

62 Vowel Counter Easy strings charAt case-insensitive

Write a program that reads a line of text and counts its vowels (a, e, i, o, u), treating uppercase and lowercase the same. Also list which different vowels appeared. If there are no vowels, say so. For example, "Programming is Fun" has 5 vowels, and the vowels found are a, i, o, u.

Sample input

Programming is Fun

Sample output

Total vowels = 5
Vowels found: a i o u

Tip: Convert the text to lower case first so you only need to check five characters. String.indexOf(ch) >= 0 is a quick way to test whether ch is a vowel.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class VowelCount {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a sentence: ");
        String s = sc.nextLine().toLowerCase();

        int count = 0;
        for (int i = 0; i < s.length(); i++) {
            if ("aeiou".indexOf(s.charAt(i)) >= 0) {
                count++;
            }
        }

        if (count == 0) {
            System.out.println("No vowels found");
            return;
        }
        System.out.println("Total vowels = " + count);
        System.out.print("Vowels found:");
        for (int i = 0; i < 5; i++) {
            char v = "aeiou".charAt(i);
            if (s.indexOf(v) >= 0) {
                System.out.print(" " + v);
            }
        }
        System.out.println();
    }
}

Solved by 1 student in Fun Program Battle

63 Capital Vowels Easy strings Character methods building strings

Write a program that reads a single word, counts its vowels, and builds a new word in which every vowel is changed to upper case. For example, for "hello" the output is a vowel count of 2 and the modified word "hEllO".

Sample input

hello

Sample output

Vowels = 2
Modified word = hEllO

Tip: Strings in Java cannot be changed. Build a new String by adding one character at a time.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class CapitaliseVowels {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a word: ");
        String word = sc.next();

        int count = 0;
        String result = "";
        for (int i = 0; i < word.length(); i++) {
            char ch = word.charAt(i);
            if ("aeiouAEIOU".indexOf(ch) >= 0) {
                count++;
                result += Character.toUpperCase(ch);
            } else {
                result += ch;
            }
        }
        System.out.println("Vowels = " + count);
        System.out.println("Modified word = " + result);
    }
}

Solved by 1 student in Fun Program Battle

64 Repeat Characters Easy strings nested loops

Write a program that takes a string and an integer k, and prints the string with each character repeated k times. For example, for "abc" and 2 the output is "aabbcc". Upper and lower case are kept as they are.

Sample input

abc
2

Sample output

aabbcc

Tip: Reading a whole line with nextLine() and then a number with nextInt() works in this order. The reverse order (nextInt() then nextLine()) needs an extra nextLine() to skip the leftover newline.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class RepeatCharacters {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a string: ");
        String s = sc.nextLine();
        System.out.print("Enter repeat count: ");
        int k = sc.nextInt();

        String result = "";
        for (int i = 0; i < s.length(); i++) {
            for (int j = 0; j < k; j++) {
                result += s.charAt(i);
            }
        }
        System.out.println(result);
    }
}

Solved by 2 students in Fun Program Battle

65 Palindrome Sentence Medium strings Character.isLetterOrDigit two pointers

Write a program that reads a string and checks whether it is a palindrome, ignoring spaces, punctuation and the difference between upper and lower case. For example, "Madam, I'm Adam" is a palindrome, but "Hello" is not.

Sample input

Madam, I'm Adam

Sample output

Palindrome

Tip: First "clean" the text by keeping only letters and digits in lower case, then compare it with its reverse. Cleaning first makes the check itself very simple.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class PalindromeSentence {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a string: ");
        String s = sc.nextLine();

        String clean = "";
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            if (Character.isLetterOrDigit(ch)) {
                clean += Character.toLowerCase(ch);    // keep letters/digits only
            }
        }

        boolean palindrome = true;
        int len = clean.length();
        for (int i = 0; i < len / 2; i++) {
            if (clean.charAt(i) != clean.charAt(len - 1 - i)) {
                palindrome = false;
                break;
            }
        }
        System.out.println(palindrome ? "Palindrome" : "Not a palindrome");
    }
}

Solved by 2 students in Fun Program Battle

66 Reverse Words Medium strings split arrays of strings

Write a program that reads a sentence and prints its words in reverse order, keeping each word itself unchanged. Extra spaces between words, and spaces at the start or end, should be ignored. For example, "Hello World from Java" becomes "Java from World Hello".

Sample input

Hello World from Java

Sample output

Java from World Hello

Tip: trim() removes the outer spaces, and split("\\s+") splits on one or more spaces, so repeated spaces never produce empty words.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class ReverseWords {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a sentence: ");
        String[] words = sc.nextLine().trim().split("\\s+");

        String result = "";
        for (int i = words.length - 1; i >= 0; i--) {
            result += words[i];
            if (i > 0) {
                result += " ";
            }
        }
        System.out.println(result);
    }
}

Solved by 2 students in Fun Program Battle

67 Chunk Reverse Medium strings substring loops with steps

Write a program that reads a string and an integer n, splits the string into chunks of n characters, reverses each chunk, and joins them back together. The last chunk may be shorter. For example, "abcdef" with n = 2 becomes "badcfe".

Sample input

abcdef
2

Sample output

badcfe

Tip: Math.min(start + n, length) stops the last chunk from going past the end of the string.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class ChunkReverse {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a string: ");
        String s = sc.nextLine();
        System.out.print("Enter chunk size: ");
        int n = sc.nextInt();

        String result = "";
        for (int start = 0; start < s.length(); start += n) {
            int end = Math.min(start + n, s.length());   // last chunk may be shorter
            for (int i = end - 1; i >= start; i--) {
                result += s.charAt(i);
            }
        }
        System.out.println(result);
    }
}

Solved by 1 student in Fun Program Battle

68 Remove Repeats Easy strings comparing neighbours

Write a program that reads a string and removes consecutive duplicate characters, so that a run of the same character becomes a single character. Case and special characters are kept. For example, "Hellooo World!!" becomes "Helo World!".

Sample input

Hellooo World!!

Sample output

Helo World!

Tip: Compare each character with the one just before it, and keep it only if it is different.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class RemoveRepeats {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a string: ");
        String s = sc.nextLine();

        String result = "";
        for (int i = 0; i < s.length(); i++) {
            if (i == 0 || s.charAt(i) != s.charAt(i - 1)) {
                result += s.charAt(i);
            }
        }
        System.out.println(result);
    }
}

Solved by 1 student in Fun Program Battle

69 Unique Letters Medium strings counting array two passes

Write a program that reads a string and finds the letters that appear exactly once, ignoring case and anything that is not a letter. Print those letters in the order they appear, and their count. If there are none, print "No unique characters found". For example, in "Hello, World!" the letters h, e, w, r and d appear once, so the count is 5.

Sample input

Hello, World!

Sample output

Unique letters: h e w r d
Count = 5

Tip: An int[26] array counts every letter, using ch - 'a' as the index. A second pass over the string then prints the letters whose count is 1, in their original order.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class UniqueLetters {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a string: ");
        String s = sc.nextLine().toLowerCase();

        int[] count = new int[26];
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            if (ch >= 'a' && ch <= 'z') {
                count[ch - 'a']++;              // 'a' -> 0, 'b' -> 1, ...
            }
        }

        String found = "";
        int unique = 0;
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            if (ch >= 'a' && ch <= 'z' && count[ch - 'a'] == 1) {
                found += " " + ch;
                unique++;
            }
        }

        if (unique == 0) {
            System.out.println("No unique characters found");
        } else {
            System.out.println("Unique letters:" + found);
            System.out.println("Count = " + unique);
        }
    }
}

Solved by 2 students in Fun Program Battle

70 Anagram Checker Medium strings counting array

Two strings are anagrams if they contain exactly the same letters the same number of times, just in a different order. Write a program that reads two strings and checks whether they are anagrams, ignoring spaces and case. For example, "Listen" and "Silent" are anagrams, and so are "Dormitory" and "Dirty room".

Sample input

Listen
Silent

Sample output

Anagrams

Tip: Instead of sorting, count the letters: add 1 for every letter in the first string and subtract 1 for every letter in the second. They are anagrams only if every count ends at 0.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class AnagramCheck {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter first string: ");
        String a = sc.nextLine().toLowerCase();
        System.out.print("Enter second string: ");
        String b = sc.nextLine().toLowerCase();

        int[] count = new int[26];
        for (int i = 0; i < a.length(); i++) {
            char ch = a.charAt(i);
            if (ch >= 'a' && ch <= 'z') {
                count[ch - 'a']++;
            }
        }
        for (int i = 0; i < b.length(); i++) {
            char ch = b.charAt(i);
            if (ch >= 'a' && ch <= 'z') {
                count[ch - 'a']--;
            }
        }

        boolean anagram = true;
        for (int i = 0; i < 26; i++) {
            if (count[i] != 0) {
                anagram = false;
                break;
            }
        }
        System.out.println(anagram ? "Anagrams" : "Not anagrams");
    }
}

Solved by 6 students in Fun Program Battle

71 Word Chain Easy strings charAt concatenation

Write a program that reads two words and checks whether the last character of the first word is the same as the first character of the second word. If it is, print the two words joined together. Otherwise print a message saying they cannot be chained. For example, "apple" and "eagle" can be chained to give "appleeagle".

Sample input

apple
eagle

Sample output

Chained: appleeagle

Tip: The last character of a string s is s.charAt(s.length() - 1). Forgetting the "- 1" is one of the most common String errors.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class ChainWords {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter first word: ");
        String a = sc.next();
        System.out.print("Enter second word: ");
        String b = sc.next();

        char last = a.charAt(a.length() - 1);     // index of the last character is length - 1
        char first = b.charAt(0);

        if (last == first) {
            System.out.println("Chained: " + a + b);
        } else {
            System.out.println("These words cannot be chained");
        }
    }
}

Solved by 1 student in Fun Program Battle

Topic 7 · 4 problems

Matrices

2-D arrays: diagonals, boundaries and mirror images.

72 Diagonal Sums Easy 2-D arrays diagonals

Write a program that reads a square matrix of size n x n and prints the sum of its main diagonal (top-left to bottom-right) and the sum of its secondary diagonal (top-right to bottom-left). For example, for the matrix 1 2 3 / 4 5 6 / 7 8 9 both sums are 15 (1 + 5 + 9 and 3 + 5 + 7).

Sample input

Sample output

Main diagonal sum = 15
Secondary diagonal sum = 15

Tip: On the main diagonal the row equals the column (a[i][i]). On the secondary diagonal the column is n - 1 - i. One loop handles both.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class DiagonalSums {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter size of matrix: ");
        int n = sc.nextInt();
        int[][] a = new int[n][n];
        System.out.print("Enter the elements row by row: ");
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                a[i][j] = sc.nextInt();
            }
        }

        int main = 0, secondary = 0;
        for (int i = 0; i < n; i++) {
            main += a[i][i];                  // row == column
            secondary += a[i][n - 1 - i];     // row + column == n - 1
        }
        System.out.println("Main diagonal sum = " + main);
        System.out.println("Secondary diagonal sum = " + secondary);
    }
}

Solved by 5 students in Fun Program Battle

73 Boundary Sum Easy 2-D arrays boundary conditions

Write a program that reads the number of rows and columns of a matrix, then its elements, and prints the sum of the elements on its boundary: the first row, the last row, the first column and the last column. It must also work when the matrix has only one row or one column. For example, for 1 2 3 / 4 5 6 / 7 8 9 the boundary sum is 40 (everything except the 5 in the middle).

Sample input

Sample output

Boundary sum = 40

Tip: Checking each cell with one condition (i == 0 || i == r - 1 || j == 0 || j == c - 1) never counts a corner twice. Adding the four edges separately would.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class BoundarySum {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter rows and columns: ");
        int r = sc.nextInt();
        int c = sc.nextInt();
        int[][] a = new int[r][c];
        System.out.print("Enter the elements row by row: ");
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                a[i][j] = sc.nextInt();
            }
        }

        int sum = 0;
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                if (i == 0 || i == r - 1 || j == 0 || j == c - 1) {
                    sum += a[i][j];           // each boundary cell is added once
                }
            }
        }
        System.out.println("Boundary sum = " + sum);
    }
}

Solved by 1 student in Fun Program Battle

74 Diagonal Matrix Easy 2-D arrays flag variable

A diagonal matrix is a square matrix in which every element outside the main diagonal is zero. Write a program that reads the size n and an n x n matrix, and prints "Yes" if it is a diagonal matrix, otherwise "No".

Sample input

Sample output

Yes

Tip: To prove a property for every element you must check them all. To disprove it, a single counter-example is enough, so you can stop at the first non-zero element off the diagonal.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class DiagonalMatrix {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter size of matrix: ");
        int n = sc.nextInt();
        int[][] a = new int[n][n];
        System.out.print("Enter the elements row by row: ");
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                a[i][j] = sc.nextInt();
            }
        }

        boolean diagonal = true;
        for (int i = 0; i < n && diagonal; i++) {
            for (int j = 0; j < n; j++) {
                if (i != j && a[i][j] != 0) {    // off-diagonal element must be 0
                    diagonal = false;
                    break;
                }
            }
        }
        System.out.println(diagonal ? "Yes" : "No");
    }
}

Solved by 1 student in Fun Program Battle

75 Mirror Matrix Easy 2-D arrays reversing rows

Write a program that reads a matrix with r rows and c columns and prints its mirror image across the vertical axis, so that each row appears reversed. For example, the matrix 1 2 3 / 4 5 6 becomes 3 2 1 / 6 5 4.

Sample input

2 3
1 2 3
4 5 6

Sample output

3 2 1
6 5 4

Tip: Column j of the mirror image is column c - 1 - j of the original. You can print the mirror directly without creating a second matrix.

Try writing it yourself first!

Java · tested

import java.util.Scanner;

public class MirrorMatrix {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter rows and columns: ");
        int r = sc.nextInt();
        int c = sc.nextInt();
        int[][] a = new int[r][c];
        System.out.print("Enter the elements row by row: ");
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                a[i][j] = sc.nextInt();
            }
        }

        for (int i = 0; i < r; i++) {
            for (int j = c - 1; j >= 0; j--) {      // read each row backwards
                System.out.print(a[i][j] + " ");
            }
            System.out.println();
        }
    }
}

Solved by 1 student in Fun Program Battle

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