leetcode-problemset/leetcode-cn/originData/minimum-number-of-flips-to-make-the-binary-string-alternating.json

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            "questionId": "2017",
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            "categoryTitle": "Algorithms",
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            "title": "Minimum Number of Flips to Make the Binary String Alternating",
            "titleSlug": "minimum-number-of-flips-to-make-the-binary-string-alternating",
            "content": "<p>You are given a binary string <code>s</code>. You are allowed to perform two types of operations on the string in any sequence:</p>\n\n<ul>\n\t<li><strong>Type-1: Remove</strong> the character at the start of the string <code>s</code> and <strong>append</strong> it to the end of the string.</li>\n\t<li><strong>Type-2: Pick</strong> any character in <code>s</code> and <strong>flip</strong> its value, i.e., if its value is <code>&#39;0&#39;</code> it becomes <code>&#39;1&#39;</code> and vice-versa.</li>\n</ul>\n\n<p>Return <em>the <strong>minimum</strong> number of <strong>type-2</strong> operations you need to perform</em> <em>such that </em><code>s</code> <em>becomes <strong>alternating</strong>.</em></p>\n\n<p>The string is called <strong>alternating</strong> if no two adjacent characters are equal.</p>\n\n<ul>\n\t<li>For example, the strings <code>&quot;010&quot;</code> and <code>&quot;1010&quot;</code> are alternating, while the string <code>&quot;0100&quot;</code> is not.</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;111000&quot;\n<strong>Output:</strong> 2\n<strong>Explanation</strong>: Use the first operation two times to make s = &quot;100011&quot;.\nThen, use the second operation on the third and sixth elements to make s = &quot;10<u>1</u>01<u>0</u>&quot;.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;010&quot;\n<strong>Output:</strong> 0\n<strong>Explanation</strong>: The string is already alternating.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;1110&quot;\n<strong>Output:</strong> 1\n<strong>Explanation</strong>: Use the second operation on the second element to make s = &quot;1<u>0</u>10&quot;.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>\n</ul>\n",
            "translatedTitle": "使二进制字符串字符交替的最少反转次数",
            "translatedContent": "<p>给你一个二进制字符串 <code>s</code> 。你可以按任意顺序执行以下两种操作任意次：</p>\n\n<ul>\n\t<li><strong>类型 1 ：删除</strong> 字符串 <code>s</code> 的第一个字符并将它 <strong>添加</strong> 到字符串结尾。</li>\n\t<li><strong>类型 2 ：选择 </strong>字符串 <code>s</code> 中任意一个字符并将该字符 <strong>反转 </strong>，也就是如果值为 <code>'0'</code> ，则反转得到 <code>'1'</code> ，反之亦然。</li>\n</ul>\n\n<p>请你返回使 <code>s</code> 变成 <strong>交替</strong> 字符串的前提下， <strong>类型 2 </strong>的 <strong>最少</strong> 操作次数 。</p>\n\n<p>我们称一个字符串是 <strong>交替</strong> 的，需要满足任意相邻字符都不同。</p>\n\n<ul>\n\t<li>比方说，字符串 <code>\"010\"</code> 和 <code>\"1010\"</code> 都是交替的，但是字符串 <code>\"0100\"</code> 不是。</li>\n</ul>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>s = \"111000\"\n<b>输出：</b>2\n<b>解释：</b>执行第一种操作两次，得到 s = \"100011\" 。\n然后对第三个和第六个字符执行第二种操作，得到 s = \"10<strong>1</strong>01<strong>0</strong>\" 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>s = \"010\"\n<b>输出：</b>0\n<strong>解释：</strong>字符串已经是交替的。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><b>输入：</b>s = \"1110\"\n<b>输出：</b>1\n<b>解释：</b>对第二个字符执行第二种操作，得到 s = \"1<strong>0</strong>10\" 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>s[i]</code> 要么是 <code>'0'</code> ，要么是 <code>'1'</code> 。</li>\n</ul>\n",
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                "Note what actually matters is how many 0s and 1s are in odd and even positions",
                "For every cyclic shift we need to count how many 0s and 1s are at each parity and convert the minimum between them for each parity"
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