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leetcode-problemset/leetcode-cn/originData/minimum-flips-to-make-a-or-b-equal-to-c.json
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{
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"question": {
"questionId": "1441",
"questionFrontendId": "1318",
"categoryTitle": "Algorithms",
"boundTopicId": 70115,
"title": "Minimum Flips to Make a OR b Equal to c",
"titleSlug": "minimum-flips-to-make-a-or-b-equal-to-c",
"content": "<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 />\r\nFlip 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>\r\n\r\n<p>&nbsp;</p>\r\n<p><strong class=\"example\">Example 1:</strong></p>\r\n\r\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/01/06/sample_3_1676.png\" style=\"width: 260px; height: 87px;\" /></p>\r\n\r\n<pre>\r\n<strong>Input:</strong> a = 2, b = 6, c = 5\r\n<strong>Output:</strong> 3\r\n<strong>Explanation: </strong>After flips a = 1 , b = 4 , c = 5 such that (<code>a</code> OR <code>b</code> == <code>c</code>)</pre>\r\n\r\n<p><strong class=\"example\">Example 2:</strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong> a = 4, b = 2, c = 7\r\n<strong>Output:</strong> 1\r\n</pre>\r\n\r\n<p><strong class=\"example\">Example 3:</strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong> a = 1, b = 2, c = 3\r\n<strong>Output:</strong> 0\r\n</pre>\r\n\r\n<p>&nbsp;</p>\r\n<p><strong>Constraints:</strong></p>\r\n\r\n<ul>\r\n\t<li><code>1 &lt;= a &lt;= 10^9</code></li>\r\n\t<li><code>1 &lt;= b&nbsp;&lt;= 10^9</code></li>\r\n\t<li><code>1 &lt;= c&nbsp;&lt;= 10^9</code></li>\r\n</ul>",
"translatedTitle": "或运算的最小翻转次数",
"translatedContent": "<p>给你三个正整数&nbsp;<code>a</code>、<code>b</code> 和 <code>c</code>。</p>\n\n<p>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。</p>\n\n<p>「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 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/01/11/sample_3_1676.png\" style=\"height: 87px; width: 260px;\"></p>\n\n<pre><strong>输入:</strong>a = 2, b = 6, c = 5\n<strong>输出:</strong>3\n<strong>解释:</strong>翻转后 a = 1 , b = 4 , c = 5 使得 <code>a</code> OR <code>b</code> == <code>c</code></pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre><strong>输入:</strong>a = 4, b = 2, c = 7\n<strong>输出:</strong>1\n</pre>\n\n<p><strong>示例 3</strong></p>\n\n<pre><strong>输入:</strong>a = 1, b = 2, c = 3\n<strong>输出:</strong>0\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= a &lt;= 10^9</code></li>\n\t<li><code>1 &lt;= b&nbsp;&lt;= 10^9</code></li>\n\t<li><code>1 &lt;= c&nbsp;&lt;= 10^9</code></li>\n</ul>\n",
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"Check the bits one by one whether they need to be flipped."
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