# 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

• 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;
}