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

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

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

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

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

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

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

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

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

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code> only contains lower case English characters and parentheses.</li>
	<li>It is guaranteed that all parentheses are balanced.</li>
</ul>

class Solution:
    def reverseParentheses(self, s: str) -> str:
        seen = []

        for val in s:
            if val == ')':
                temp = []
                while seen[-1] != '(':
                    temp.append(seen.pop())
                seen.pop()
                seen.extend(temp)
            else:
                seen.append(val)
        
        return ''.join(seen)        
16500