Oddly even– TCS NQT Question 2

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 4 and 6 as they are pos: 1 and pos: 3, both have sum 10. Similarly, 5 and 7 are at even positions pos: 2 and pos: 4 with sum 12. Thus, difference is 12 – 10 = 2

Case 2

• Input: 5476
• Expected Output: 2

Case 3

• Input: 9834698765123
• Expected Output: 1

num = [int(d) for d in str(input())]
even,odd = 0,0
for i in range(0,len(num)):
    if i % 2 ==0:
        even = even + num[i]
    else:
        odd = odd + num[i]
print(abs(odd-even))

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