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:

<li><code>1 &lt;= a &lt;= 10^9</code></li>

<li><code>1 &lt;= b&nbsp;&lt;= 10^9</code></li>

<li><code>1 &lt;= c&nbsp;&lt;= 10^9</code></li>

Solutions

Python3

class Solution:
    def minFlips(self, a: int, b: int, c: int) -> int:
        ans = 0
        for i in range(31):
            x, y, z = (a >> i) & 1, (b >> i) & 1, (c >> i) & 1
            if (x | y) == z:
                continue
            if x == 1 and y == 1 and z == 0:
                ans += 2
            else:
                ans += 1
        return ans

Java

class Solution {
    public int minFlips(int a, int b, int c) {
        int ans = 0;
        for (int i = 0; i < 31; ++i) {
            int x = (a >> i) & 1, y = (b >> i) & 1, z = (c >> i) & 1;
            if ((x | y) == z) {
                continue;
            }
            if (x == 1 && y == 1 && z == 0) {
                ++ans;
            }
            ++ans;
        }
        return ans;
    }
}

C++

class Solution {
public:
    int minFlips(int a, int b, int c) {
        int ans = 0;
        for (int i = 0; i < 31; ++i) {
            int x = (a >> i) & 1, y = (b >> i) & 1, z = (c >> i) & 1;
            if ((x | y) == z) continue;
            if (x == 1 && y == 1 && z == 0) ++ans;
            ++ans;
        }
        return ans;
    }
};

Go

func minFlips(a int, b int, c int) int {
	ans := 0
	for i := 0; i < 31; i++ {
		x, y, z := (a>>i)&1, (b>>i)&1, (c>>i)&1
		if (x | y) == z {
			continue
		}
		if x == 1 && y == 1 && z == 0 {
			ans++
		}
		ans++
	}
	return ans
}

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