Minimum Flips to Make a OR b Equal to c in C++

Suppose we have 3 positives numbers a, b and c. We have to find the minimum flips required in some bits of a and b to make (a OR b == c ). Here we are considering bitwise OR operation.

The flip operation consists of change any single bit 1 to 0 or change the bit 0 to 1 in their binary representation. So if a : 0010 and b := 0110, so c is 0101, After flips, a will be 0001, and b will be 0100

To solve this, we will follow these steps −

ans := 0
for i in range 0 to 31
- bitC := (c / 2^i) AND 1
- bitA := (a / 2^i) AND 1
- bitB := (b / 2^i) AND 1
- if (bitA OR bitB) is not same as bitC, then
  - if bitC is 0
    - if bitA = 1 and bitB = 1, then increase ans by 2, otherwise increase ans by 1
  - otherwise increase ans by 1
return ans

Example(C++)

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   int minFlips(int a, int b, int c) {
      int ans = 0;
      for(int i = 0; i < 32; i++){
         int bitC = (c >> i) & 1;
         int bitA = (a >> i) & 1;
         int bitB = (b >> i) & 1;
         if((bitA || bitB) != bitC){
            if(!bitC){
               if(bitA == 1 && bitB == 1){
                  ans += 2;
               }
               else {
                  ans += 1;
               }
            }
            else{
               ans += 1;
            }
         }
      }
      return ans;
   }
};
main(){
   Solution ob;
   cout << (ob.minFlips(2,6,5));
}