Example 1:

Input: a = 2, b = 6, c = 5
Output: 3
Explanation: After flips a = 1 , b = 4 , c = 5 such that (a OR b == c)

Example 2:

Input: a = 4, b = 2, c = 7
Output: 1

Example 3:

Input: a = 1, b = 2, c = 3
Output: 0

Constraints:

• 1 <= a <= 10^9
• 1 <= b <= 10^9
• 1 <= c <= 10^9

Solution: Bit operation

If the bit of c is 1, a / b at least has one 1. cost = 1 – ((a | b) & 1)
If the bit of c is 0, a / b must be 0, cost = (a & 1) + (b & 1)

Time complexity: O(32)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int minFlips(int a, int b, int c) {
    int count = 0;
    for (int i = 0; i < 32; ++i) {      
      if (c & 1) count += 1 - ((a | b) & 1);
      else count += (a & 1) + (b & 1);
      a >>= 1;
      b >>= 1;
      c >>= 1;
    }
    return count;
  }
};

The post 花花酱 LeetCode 1318. Minimum Flips to Make a OR b Equal to c appeared first on Huahua's Tech Road.