<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) {
        int n = s.length();
        stack<int> openParenthesesIndices;
        vector<int> pair(n);

        // First pass: Pair up parentheses
        for (int i = 0; i < n; ++i) {
            if (s[i] == '(') {
                openParenthesesIndices.push(i);
            }
            if (s[i] == ')') {
                int j = openParenthesesIndices.top();
                openParenthesesIndices.pop();
                pair[i] = j;
                pair[j] = i;
            }
        }

        // Second pass: Build the result string
        string result;
        for (int currIndex = 0, direction = 1; currIndex < n;
             currIndex += direction) {
            if (s[currIndex] == '(' || s[currIndex] == ')') {
                currIndex = pair[currIndex];
                direction = -direction;
            } else {
                result += s[currIndex];
            }
        }
        return result;
    }
};
8100