<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 {
public:
    string reverseParentheses(string s) {
        stack<int> forward;

        for (int i = 0; i < s.size(); i++) {
            //cout << s[i] << endl;
            if (s[i] == '(') {
                forward.push(i);
            }
            else if (s[i] == ')') {
                int start = s.size();

                string portion = s.substr(forward.top() + 1, i - forward.top() - 1);
                //cout << portion << endl;
                s.replace(forward.top(), i - forward.top() + 1, string(portion.rbegin(), portion.rend()));
                forward.pop();

                //cout << i << " -= " << start << " - " << s.size() << endl;
                i -= start - s.size();
            }
            
        }

        return s;
    }
};
9000