<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 {
    bool CheckValid(string &str)
    {
        stack<char> st;
        int i = 0, n = str.size();
        while(i < n)
        {
            if(str[i] == '(')
            {
                st.push(str[i]);
            }
            else if(str[i] == ')')
            {
                if(st.empty())
                    return false;
                st.pop();
            }
            i++;
        }
        if(st.empty())
            return true;
        return false;
    }
public:
    vector<string> removeInvalidParentheses(string s) {
        int n = s.size();
        queue<string> q;
        q.push(s);
        vector<string> um;
        unordered_map<string, int> visited;
        bool bk = false;
        if(CheckValid(s))
        {
            um.push_back(s);
            return um;
        }
        while(!q.empty())
        {
            int sz = q.size();
            while(sz--)
            {
                string tmp = q.front();
                q.pop();
                for(int i = 0; i < tmp.size(); i++)
                {
                    if(s[i] >= 'a' && s[i] <= 'z') continue;
                    string z = tmp;
                    z.erase(i, 1);
                    if(visited.count(z)) continue;
                    if(CheckValid(z))
                    {
                        um.push_back(z);
                        bk = true;
                    }
                    visited[z]++;
                    q.push(z);
                }
            }
            if(bk) break;
        }
        return um;
    }
};
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