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1318. Minimum Flips to Make a OR b Equal to c

Description

Given 3 positives numbers a, b and c. Return the minimum flips required in some bits of a and b to make ( a OR b == c ). (bitwise OR operation).
Flip operation consists of change any single bit 1 to 0 or change the bit 0 to 1 in their binary representation.

 

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

Solutions

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public int minFlips(int a, int b, int c) {
          int ans = 0;
          for (int i = 0; i < 30; ++i) {
              int x = a >> i & 1, y = b >> i & 1, z = c >> i & 1;
              if ((x | y) != z) {
                  ans += x == 1 && y == 1 ? 2 : 1;
              }
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int minFlips(int a, int b, int c) {
          int ans = 0;
          for (int i = 0; i < 30; ++i) {
              int x = a >> i & 1, y = b >> i & 1, z = c >> i & 1;
              if ((x | y) != z) {
                  ans += x == 1 && y == 1 ? 2 : 1;
              }
          }
          return ans;
      }
  };
  

• class Solution:
      def minFlips(self, a: int, b: int, c: int) -> int:
          ans = 0
          for i in range(30):
              x, y, z = a >> i & 1, b >> i & 1, c >> i & 1
              if x | y != z:
                  ans += 2 if x == 1 and y == 1 else 1
          return ans
  
  

• func minFlips(a int, b int, c int) (ans int) {
  	for i := 0; i < 30; i++ {
  		x, y, z := a>>i&1, b>>i&1, c>>i&1
  		if (x | y) != z {
  			if x == 1 && y == 1 {
  				ans += 2
  			} else {
  				ans++
  			}
  		}
  	}
  	return
  }
  

• function minFlips(a: number, b: number, c: number): number {
      let ans = 0;
      for (let i = 0; i < 32; ++i) {
          const [x, y, z] = [(a >> i) & 1, (b >> i) & 1, (c >> i) & 1];
          ans += z === 0 ? x + y : x + y === 0 ? 1 : 0;
      }
      return ans;
  }
  
  

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