README_EN.md

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

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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

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

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