TCS Coding Question - Even Sum and Odd Sum Difference | Given a maximum of 100 digit numbers as input, find the difference between the sum of odd and even position digits

 TCS Coding Question - Odd Even Place Sum Difference

Problem Statement:

Given a maximum of 100 digit numbers as input, find the difference between the sum of odd and even position digits

Test Cases

Case 1

Input: 4567

Expected Output: 2

Explanation : Odd positions are pos: 1 and pos: 3 which has digits 4 and 6 respectively, and both have sum 10. Similarly, even positions pos: 2 and pos: 4  with digits 5 and 7 respectively and sum 12. Thus, difference is 12 – 10 = 2

Case 2

Input: 5476

Expected Output: 2

Case 3

Input: 9834698765123

Expected Output: 1

Approach:

1. Calculate sum for digits at Odd places

2. Calculate sum for digits at even places

3. Find difference between both the Sum

4. Return result

C Progam

#include<stdio.h>
#include<string.h>
#include <stdlib.h>
int main()
{
    int oddSum = 0,evenSum = 0,i = 0, n,diff;
    long long num;
    scanf("%lld",&num);  
    
    while(num != 0){
        if(i%2==0){
            evenSum = evenSum + num%10;
            num = num/10;
            i++;
        }
        else{
            oddSum = oddSum + num%10;
            num = num/10;
            i++;
        }
    }
    printf("%d",abs(oddSum - evenSum));
    return 0;
}

Java Program

package codingExample;

import java.util.Scanner;

public class OddEvenPositionSumDifference {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		String input = sc.nextLine();

		long evenSum = 0, oddSum = 0;
		
		for (int i = 0; i < input.length(); i++) {
			if (i % 2 == 0)
				evenSum = evenSum + input.charAt(i) - '0';
			else
				oddSum = oddSum + input.charAt(i) - '0';
		}
		System.out.println(Math.abs(evenSum - oddSum));
	}
}

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