leetcode-problemset/leetcode-cn/problem (English)/或运算的最小翻转次数(English) [minimum-flips-to-make-a-or-b-equal-to-c].html

<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://assets.leetcode.com/uploads/2020/01/06/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>