<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_set<string> setAnsr;
public:
    void removeParenthensis(string &s,int ind,int leftP,int rightP,string tmp,int rc){
        if(ind==s.size()){
            if(rc==0 && leftP==0 && rightP==0)
            setAnsr.insert(tmp);
            return ;
        }
        if(rc<0||leftP<0||rightP<0)
        return ;
        if(s[ind]=='(')
            removeParenthensis(s,ind+1,leftP-1,rightP,tmp,rc);
        else if(s[ind]==')')
            removeParenthensis(s,ind+1,leftP,rightP-1,tmp,rc);
        if(s[ind]=='(')
        rc++;
        if(s[ind]==')')
        rc--;
        removeParenthensis(s,ind+1,leftP,rightP,tmp+s[ind],rc);
    }
    vector<string> removeInvalidParentheses(string s) {
        int leftP=0,rightP=0;
        // get minimum of these needed
        // call the function to calculate answer
        int tc=0;
        for(int i=0;i<s.size();++i){
            if(s[i]=='(')
            tc++;
            else if(s[i]==')')
            tc--;
            if(tc<0){
                rightP++;
                tc=0;
            }
        }
        if(tc>0)
        leftP=tc;
        removeParenthensis(s,0,leftP,rightP,"",0);
        vector<string> ansr;
        for(auto it=setAnsr.begin();it!=setAnsr.end();++it)
        ansr.push_back(*it);
        return ansr;
    }
};
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