<p>Given a string <code>s</code> that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.</p>

<p>Return <em>a list of <strong>unique strings</strong> that are valid with the minimum number of removals</em>. You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;()())()&quot;
<strong>Output:</strong> [&quot;(())()&quot;,&quot;()()()&quot;]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;(a)())()&quot;
<strong>Output:</strong> [&quot;(a())()&quot;,&quot;(a)()()&quot;]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;)(&quot;
<strong>Output:</strong> [&quot;&quot;]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 25</code></li>
	<li><code>s</code> consists of lowercase English letters and parentheses <code>&#39;(&#39;</code> and <code>&#39;)&#39;</code>.</li>
	<li>There will be at most <code>20</code> parentheses in <code>s</code>.</li>
</ul>

class Solution {
    unordered_map<string,bool> dp;
    int validatePrenthesis(string &s){
       int n = s.size();
       stack<char> st;
       int count = 0;
       for(int i=0;i<n;i++){
          if(s[i] == '('){
             st.push('(');
          }else if(s[i] == ')'){
             if(!st.empty() && st.top() == '('){
                st.pop();
             }else{
                count++;
             }
          }
       }
       count += st.size();
       return count;
    }
    void solve(int index,string &s,set<string> &ans,string temp,int limit){
        if(index == s.size()){
            if(limit == 0 && validatePrenthesis(temp) == 0){
                ans.insert(temp);
            }
            return;
        }
        string key = to_string(index)+"."+temp+"."+to_string(limit);
        if(dp.count(key)){
            return;
        }
        if(s[index] == '(' || s[index] == ')'){
            solve(index+1,s,ans,temp+s[index],limit);
            solve(index+1,s,ans,temp,limit-1);
        }else{
            solve(index+1,s,ans,temp+s[index],limit);
        }
        dp[key] = true;
    }
public:
    vector<string> removeInvalidParentheses(string s) {
       int count = validatePrenthesis(s);
       cout<<count<<endl;
       if(count == 0){
          return {s};
       }
       set<string> ans;
       string temp = "";
       solve(0,s,ans,temp,count);
       vector<string> res (ans.begin(),ans.end());
       return res;
    }
};
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