## 13377 - bbits

## Brief
find the minimum flips required in 3 positive integers x,y,z. Such that (X OR Y == Z), a bitwise OR operation.

## Input
Input is three positive integers, x, y, z.

0 < x,y,z < 109
## Output
Output is minimum number of flips to make x OR y == z

Don't forget to include "\n"
## Solution

//by Mary Madeline
#include<stdio.h>

int minFlips(int a, int b, int c){
    int difference = (a | b) ^ c;
    int num = difference;
    int i= 0;
    int flips = 0;

    while (num){
        /* calculate only for misplaced bits */
        if(difference & (1<<i)){
            // case for 1 condition and 0 condition
            if(c & (1<<i)){
                flips += 1;
            }
            else{
                if(a & (1<<i)) flips += 1;
                if(b & (1<<i)) flips += 1;   
            } 
        }  
        i += 1;
        num >>= 1;
    }

return flips;
}


int main(){
    int x,y,z;
    scanf("%d %d %d", &x, &y, &z);
    printf("%d\n", minFlips(x,y,z));
}