Minimum Flips to Make a OR b Equal to c
<p>Given 3 positives numbers <code>a</code>, <code>b</code> and <code>c</code>. Return the minimum flips required in some bits of <code>a</code> and <code>b</code> to make (&nbsp;<code>a</code> OR <code>b</code> == <code>c</code>&nbsp;). (bitwise OR operation).<br />

Flip operation&nbsp;consists of change&nbsp;<strong>any</strong>&nbsp;single bit 1 to 0 or change the bit 0 to 1&nbsp;in their binary representation.</p>

<p>&nbsp;</p>

<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1318.Minimum%20Flips%20to%20Make%20a%20OR%20b%20Equal%20to%20c/images/sample_3_1676.png" style="width: 260px; height: 87px;" /></p>

<pre>

<strong>Input:</strong> a = 2, b = 6, c = 5

<strong>Output:</strong> 3

<strong>Explanation: </strong>After flips a = 1 , b = 4 , c = 5 such that (<code>a</code> OR <code>b</code> == <code>c</code>)</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>

<strong>Input:</strong> a = 4, b = 2, c = 7

<strong>Output:</strong> 1

</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>

<strong>Input:</strong> a = 1, b = 2, c = 3

<strong>Output:</strong> 0

</pre>

<p>&nbsp;</p>

<p><strong>Constraints:</strong></p>

<ul>

    <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>

</ul>
Bit Manipulation
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
}