Want to know what is wrong in this code and explain how it can be corrected

#1

#include
using namespace std;
int main() {
long long int x;
cin>>x;
long long int v=x;
long long int sum= 0;
long long int mul= 1;
while(v!=0){
long long int rem =v%10;
v=v/10;
if(rem>=5){
rem=9-rem;
sum=sum+remmul;
}
else if(rem==9 && v==0){
sum=sum+9mul;
}
else{
sum=sum+remmul;
}
mul=mul10;
}
cout<<sum;
}

#2

hi @jharshitjindal_4da85e4d0b87909f
if digit is >5 flip that digit otherwise don’t flip
also check some corner cases
a) first digit should not be zero so if first digit is 9 then it should remain unchanged
b) if number is 0 only then output should be 9, not 0.

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 in c++ and long in java.

for implementation difficulties refer https://ide.codingblocks.com/s/658154

#3

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