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

## LeetCode [1318. Minimum Flips to Make a OR b Equal to c](https://github.com/txfs19260817/LeetCode-Go/blob/master/solutions/title/README.md)

### 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:**

![](https://assets.leetcode.com/uploads/2020/01/06/sample_3_1676.png)

```
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`

### Tags

Bit Manipulation

### Solution

Check the bits one by one (via `(x>>i)&1`), and:

* if `c[i] == 1` and `a[i] == b[i] == 0`, then increase the counter by 1;
* if `c[i] == 0` and any of `a[i] == 1` or `b[i] == 1` will let the counter to increment itself by 1.

### Complexity

* Time complexity: $$O(1)$$, no greater than 32 times;
* Space complexity: $$O(1)$$

### Code

```go
func minFlips(a int, b int, c int) int {
	var ans int
	for i := 0; i < 31; i++ {
		if (c>>i)&1 == 1 && (a>>i)&1 == 0 && (b>>i)&1 == 0 {
			ans++
		} else if (c>>i)&1 == 0 {
			if (a>>i)&1 == 1 {
				ans++
			}
			if (b>>i)&1 == 1 {
				ans++
			}
		}
	}
	return ans
}
```