<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 {
public:
bool ck(string s){
    stack<char>st;
    for(int i=0;i<s.size();i++){
        if(s[i]=='('){
            st.push('(');
        }
        else if(s[i]==')'){
            if(st.empty())return false;
            st.pop();
        }
    }
    if(st.empty())return true;
    return false;
}
    vector<string> removeInvalidParentheses(string s) {
        vector<string>ans;
        queue<string>q;
        set<string>vis;
        int n=s.size();
        q.push(s);
        if(ck(s)){
            ans.push_back(s);
            return ans;
        }
        while(!q.empty()){
            int sz=q.size();
            bool ok=false;
            while(sz--){
                string t=q.front();
                q.pop();
                for(int i=0;i<t.size();i++){
               
                if(s[i]>='a' && s[i]<='z')continue;
                 string temp=t;
                temp.erase(i,1);
                    if(vis.count(temp))continue;
                    if(ck(temp)){
                        ok=true;
                        ans.push_back(temp);
                    }
                    vis.insert(temp);
                    q.push(temp);
                }
            }
            if(ok)break;
        }
        return ans;

    }
};
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