Java Code Example: Calculate Cross Sum (iterative and recursive)

The cross sum (or digit sum) of a number is the sum of its individual digits. For example, the cross sum of 1234 is 1 + 2 + 3 + 4 = 10. This concept is often used in mathematical algorithms and checksums. In this guide, we will explore two approaches to calculating the cross sum in Java: an iterative approach using loops and a recursive approach using function recursion.

Iterative Approach

In the iterative approach, we repeatedly extract the last digit of the number using the modulus operator %, add it to the sum, and remove the last digit using integer division / until the number becomes zero.

Code Example:

public class CrossSumCalculator {
    // Iterative approach to calculate the cross sum
    public static int crossSumIterative(int number) {
        int sum = 0;
        while (number != 0) {
            sum += number % 10;  // Extract the last digit and add it to sum
            number /= 10;  // Remove the last digit
        }
        return sum;
    }

    public static void main(String[] args) {
        int number = 1234;
        System.out.println("Cross sum (Iterative) of " + number + " is: " + crossSumIterative(number));
    }
}

Output:

Cross sum (Iterative) of 1234 is: 10

Explanation:

1. Initialization: We initialize sum to 0.
2. Loop Execution: We iterate while number is not 0:
   • Extract the last digit using number % 10.
   • Add this digit to sum.
   • Remove the last digit using number / 10.
3. Termination: The loop continues until number becomes 0, at which point sum contains the total cross sum.

Recursive Approach

In the recursive approach, we break the problem into smaller subproblems:

• Base case: If number == 0, return 0.
• Recursive case: Extract the last digit (number % 10) and add it to the result of the recursive call with number / 10.

Code Example:

public class CrossSumCalculator {
    // Recursive approach to calculate the cross sum
    public static int crossSumRecursive(int number) {
        if (number == 0) {
            return 0;  // Base case
        }
        return (number % 10) + crossSumRecursive(number / 10);  // Recursive case
    }

    public static void main(String[] args) {
        int number = 1234;
        System.out.println("Cross sum (Recursive) of " + number + " is: " + crossSumRecursive(number));
    }
}

Output:

Cross sum (Recursive) of 1234 is: 10

Explanation:

1. Base Case: If the number is 0, return 0 (stopping condition).
2. Recursive Case:
   • Extract the last digit using number % 10.
   • Make a recursive call with number / 10 (removing the last digit).
   • Add the extracted digit to the result of the recursive call.
3. Termination: The recursion continues until the number reduces to 0, at which point the accumulated sum is returned.

Conclusion

Both iterative and recursive methods effectively calculate the cross sum of a number. The iterative method is more efficient in terms of memory because it does not require extra stack space for recursion. The recursive method, however, provides a more elegant and intuitive way to express the problem. The choice between the two depends on the use case and constraints of your program.