minimum-flips-to-make-a-or-b-equal-to-c

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Algorithms

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). 
Flip operation&nbsp;consists of change&nbsp;any&nbsp;single bit 1 to 0 or change the bit 0 to 1&nbsp;in their binary representation.

&nbsp;
Example 1:

<img alt="" src="https://assets.leetcode.com/uploads/2020/01/06/sample_3_1676.png" style="width: 260px; height: 87px;" />

<pre>
Input: a = 2, b = 6, c = 5
Output: 3
Explanation: After flips a = 1 , b = 4 , c = 5 such that (<code>a</code> OR <code>b</code> == <code>c</code>)</pre>

Example 2:

<pre>
Input: a = 4, b = 2, c = 7
Output: 1
</pre>

Example 3:

<pre>
Input: a = 1, b = 2, c = 3
Output: 0
</pre>

&nbsp;
Constraints:

<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>

minFlips

Minimum Flips to Make a OR b Equal to c

Minimum Bit Flips to Convert Number

minimum-bit-flips-to-convert-number

转换数字的最少位翻转次数

Bit Manipulation

给你三个正整数&nbsp;<code>a</code>、<code>b</code> 和 <code>c</code>。

你可以对 <code>a</code> 和 <code>b</code>&nbsp;的二进制表示进行位翻转操作，返回能够使按位或运算&nbsp; &nbsp;<code>a</code> OR <code>b</code> == <code>c</code>&nbsp;&nbsp;成立的最小翻转次数。

「位翻转操作」是指将一个数的二进制表示任何单个位上的 1 变成 0 或者 0 变成 1 。

&nbsp;

示例 1：

<img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/01/11/sample_3_1676.png" style="height: 87px; width: 260px;">

<pre>输入：a = 2, b = 6, c = 5
输出：3
解释：翻转后 a = 1 , b = 4 , c = 5 使得 <code>a</code> OR <code>b</code> == <code>c</code></pre>

示例 2：

<pre>输入：a = 4, b = 2, c = 7
输出：1
</pre>

示例 3：

<pre>输入：a = 1, b = 2, c = 3
输出：0
</pre>

&nbsp;

提示：

<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>