README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1300-1399/1318.Minimum%20Flips%20to%20Make%20a%20OR%20b%20Equal%20to%20c/README_EN.md	1382	Weekly Contest 171 Q2	Bit Manipulation

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

Solution 1: Bit Manipulation

We can enumerate each bit of the binary representation of $a$, $b$, and $c$, denoted as $x$, $y$, and $z$ respectively. If the bitwise OR operation result of $x$ and $y$ is different from $z$, we then check if both $x$ and $y$ are $1$. If so, we need to flip twice, otherwise, we only need to flip once. We accumulate all the required flip times.

The time complexity is $O(\log M)$, where $M$ is the maximum value of the numbers in the problem. The space complexity is $O(1)$.

Python3

class Solution:
    def minFlips(self, a: int, b: int, c: int) -> int:
        ans = 0
        for i in range(32):
            x, y, z = a >> i & 1, b >> i & 1, c >> i & 1
            ans += x + y if z == 0 else int(x == 0 and y == 0)
        return ans

Java

class Solution {
    public int minFlips(int a, int b, int c) {
        int ans = 0;
        for (int i = 0; i < 32; ++i) {
            int x = a >> i & 1, y = b >> i & 1, z = c >> i & 1;
            ans += z == 0 ? x + y : (x == 0 && y == 0 ? 1 : 0);
        }
        return ans;
    }
}

C++

class Solution {
public:
    int minFlips(int a, int b, int c) {
        int ans = 0;
        for (int i = 0; i < 32; ++i) {
            int x = a >> i & 1, y = b >> i & 1, z = c >> i & 1;
            ans += z == 0 ? x + y : (x == 0 && y == 0 ? 1 : 0);
        }
        return ans;
    }
};

Go

func minFlips(a int, b int, c int) (ans int) {
	for i := 0; i < 32; i++ {
		x, y, z := a>>i&1, b>>i&1, c>>i&1
		if z == 0 {
			ans += x + y
		} else if x == 0 && y == 0 {
			ans++
		}
	}
	return
}

TypeScript

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;
}