leetcode-problemset/leetcode-cn/problem (English)/转换字符串的最少操作次数(English) [minimum-moves-to-convert-string].html

<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>

<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>

<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>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;XXX&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> <u>XXX</u> -&gt; OOO
We select all the 3 characters and convert them in one move.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;XXOX&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> <u>XXO</u>X -&gt; O<u>OOX</u> -&gt; OOOO
We select the first 3 characters in the first move, and convert them to <code>&#39;O&#39;</code>.
Then we select the last 3 characters and convert them so that the final string contains all <code>&#39;O&#39;</code>s.</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;OOOO&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no <code>&#39;X&#39;s</code> in <code>s</code> to convert.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s[i]</code> is either <code>&#39;X&#39;</code> or <code>&#39;O&#39;</code>.</li>
</ul>