leetcode-problemset/leetcode-cn/originData/minimum-moves-to-convert-string.json

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            "title": "Minimum Moves to Convert String",
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            "content": "<p>You are given a string <code>s</code> consisting of <code>n</code> characters which are either <code>&#39;X&#39;</code> or <code>&#39;O&#39;</code>.</p>\n\n<p>A <strong>move</strong> is defined as selecting <strong>three</strong> <strong>consecutive characters</strong> of <code>s</code> and converting them to <code>&#39;O&#39;</code>. Note that if a move is applied to the character <code>&#39;O&#39;</code>, it will stay the <strong>same</strong>.</p>\n\n<p>Return <em>the <strong>minimum</strong> number of moves required so that all the characters of </em><code>s</code><em> are converted to </em><code>&#39;O&#39;</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;XXX&quot;\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> <u>XXX</u> -&gt; OOO\nWe select all the 3 characters and convert them in one move.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;XXOX&quot;\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> <u>XXO</u>X -&gt; O<u>OOX</u> -&gt; OOOO\nWe select the first 3 characters in the first move, and convert them to <code>&#39;O&#39;</code>.\nThen we select the last 3 characters and convert them so that the final string contains all <code>&#39;O&#39;</code>s.</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;OOOO&quot;\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> There are no <code>&#39;X&#39;s</code> in <code>s</code> to convert.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>3 &lt;= s.length &lt;= 1000</code></li>\n\t<li><code>s[i]</code> is either <code>&#39;X&#39;</code> or <code>&#39;O&#39;</code>.</li>\n</ul>\n",
            "translatedTitle": "转换字符串的最少操作次数",
            "translatedContent": "<p>给你一个字符串 <code>s</code> ，由 <code>n</code> 个字符组成，每个字符不是 <code>'X'</code> 就是 <code>'O'</code> 。</p>\n\n<p>一次<strong> 操作</strong> 定义为从 <code>s</code> 中选出 <strong>三个连续字符 </strong>并将选中的每个字符都转换为 <code>'O'</code> 。注意，如果字符已经是 <code>'O'</code> ，只需要保持 <strong>不变</strong> 。</p>\n\n<p>返回将 <code>s</code> 中所有字符均转换为 <code>'O'</code> 需要执行的&nbsp;<strong>最少</strong>&nbsp;操作次数。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"XXX\"\n<strong>输出：</strong>1\n<strong>解释：<em>XXX</em></strong> -&gt; OOO\n一次操作，选中全部 3 个字符，并将它们转换为 <code>'O' 。</code>\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"XXOX\"\n<strong>输出：</strong>2\n<strong>解释：<em>XXO</em></strong>X -&gt; O<em><strong>OOX</strong></em> -&gt; OOOO\n第一次操作，选择前 3 个字符，并将这些字符转换为 <code>'O'</code> 。\n然后，选中后 3 个字符，并执行转换。最终得到的字符串全由字符 <code>'O'</code> 组成。</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"OOOO\"\n<strong>输出：</strong>0\n<strong>解释：</strong>s 中不存在需要转换的 <code>'X' 。</code>\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>3 &lt;= s.length &lt;= 1000</code></li>\n\t<li><code>s[i]</code> 为 <code>'X'</code> 或 <code>'O'</code></li>\n</ul>\n",
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