<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>


#define endl '\n'
#define pint pair<int, int>
#define vint vector<int>
#define vll vector<long long>
#define vpint vector<pair<int, int>>
#define vstr vector<string>
#define uset unordered_set
#define umap unordered_map
#define vbool vector<bool>
#define vvint vector<vector<int>>
#define vvvint vector<vector<vector<int>>>
#define ll long long
#define MOD 1000000007

class Solution {
public:
    string reverseParentheses(string s) {
        int n = s.length();

        stack<string> st;
        string curr;

        for (int i = 0; i < n; i++) {
            if (s[i] == '(') {
                st.push(curr);
                curr = "";
            }

            else if (s[i] >= 'a' && s[i] <= 'z')
                curr.push_back(s[i]);

            else if (s[i] == ')') {
                reverse(curr.begin(), curr.end());

                string l;
                if (!st.empty()) {
                    l = st.top();
                    st.pop();
                }

                curr = l + curr;
            }

            cout << curr << endl;
        }

        return curr;
    }
};
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