reverse-substrings-between-each-pair-of-parentheses.md

<p>You are given a string <code>s</code> that consists of lower case English letters and brackets.&nbsp;</p>

<p>Reverse the strings&nbsp;in each&nbsp;pair of matching parentheses, starting&nbsp;from the innermost one.</p>

<p>Your result should <strong>not</strong> contain any brackets.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;(abcd)&quot;
<strong>Output:</strong> &quot;dcba&quot;
</pre>

<p><strong>Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;(u(love)i)&quot;
<strong>Output:</strong> &quot;iloveu&quot;
<strong>Explanation:</strong>&nbsp;The substring &quot;love&quot; is reversed first, then the whole string is reversed.
</pre>

<p><strong>Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;(ed(et(oc))el)&quot;
<strong>Output:</strong> &quot;leetcode&quot;
<strong>Explanation:</strong>&nbsp;First, we reverse the substring &quot;oc&quot;, then &quot;etco&quot;, and finally, the whole string.
</pre>

<p><strong>Example 4:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;a(bcdefghijkl(mno)p)q&quot;
<strong>Output:</strong> &quot;apmnolkjihgfedcbq&quot;
</pre>

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

<ul>
	<li><code>0 &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code> only contains lower case English characters and parentheses.</li>
	<li>It&#39;s guaranteed that all parentheses are balanced.</li>
</ul>