<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) {
    string ans;
    stack<int> stack;
    unordered_map<int, int> pair;

    for (int i = 0; i < s.length(); ++i)
      if (s[i] == '(') {
        stack.push(i);
      } else if (s[i] == ')') {
        const int j = stack.top();
        stack.pop();
        pair[i] = j;
        pair[j] = i;
      }

    for (int i = 0, d = 1; i < s.length(); i += d)
      if (s[i] == '(' || s[i] == ')') {
        i = pair[i];
        d = -d;
      } else {
        ans += s[i];
      }

    return ans;
  }
};

static const int __  = [](){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    return 0;
}();
8400