What is the problem in the code

#1

What is the problem in my code

#2

hi @kartikstackexchange_77770317975ae8b4
The logic behind this Problem was pretty simple as we can invert any digit ( 9 - digit) but we need to invert only such digits that will eventually end up giving the smallest number possible.

So, we should invert only digits greater then or equal to 5 as after inverting them, the result gives us smallest number.

For e.g.,

9 - 5 = 4 viz, smaller than the original number that was 5.

9 - 8 = 1 viz, smaller than the original number that was 8.

but 9 - 1 = 8 viz, greater than the original number that was 1.

Important point to consider is, After inverting any digits their should not be trailing zeros that means, if their is 9 at the starting of the number then it must remain the same

Also input can be quite big so, u need to capture the number in long long int

Code -->

#include<iostream>
using namespace std;
int main() {
    long long int num, n;
    cin >> num;
    n = num;
    long long int ans = 0;
    long long int mult = 1;
    while(n != 0){

      long long int rem = n % 10;

      if(rem >= 5){

        if(rem == 9 && (n/10) == 0){
         //do nothing
        }else{
          rem = 9 - rem;
        }
      }
      ans += rem * mult;
      mult *= 10;
      n /= 10;
    }
    cout << ans;
    return 0;
}
#3

I hope I’ve cleared your doubt. I ask you to please rate your experience here
Your feedback is very important. It helps us improve our platform and hence provide you
the learning experience you deserve.

On the off chance, you still have some questions or not find the answers satisfactory, you may reopen
the doubt.