<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>st;
        map<int,int>mp;
        st.push(0);
        for(int i=1;i<n;i++){
            if(s[i] == ')'){
                mp[i] = st.top();
                mp[st.top()] =i;
                st.pop();
            }
            if(s[i] == '(') st.push(i);
        }
        string ans="";
        char flag='R';
        int i=0;
        while(i < n){
            if(s[i] == '(' || s[i] == ')'){
               i = mp[i];
               if(flag == 'R'){
                  i=i-1;
                  flag = 'L';
               }
               else if(flag == 'L'){
                  i=i+1;
                  flag ='R';
               }
            }
            else{
                ans = ans + s[i];
                if(flag == 'R') i++;
                else if(flag =='L')i--;
            }
        }
        return ans;
    }
};
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