<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 isValid(string &s){
        int open = 0;
        for(int i=0; i<s.length(); i++){
            if(s[i] == '(') open++;
            else if(s[i] == ')'){
                open--;
                if(open < 0) return false;
            }
        }
        return open == 0;
    }

    void solve(string &s, int cnt, int idx, unordered_set<string> &st){
        if(cnt < 0) return; // No more removals are possible
        if(idx == s.length()){
            if(isValid(s)){
                st.insert(s);
            }
            return;
        }
        
        if(s[idx] != '(' && s[idx] != ')'){
            solve(s, cnt, idx + 1, st);
            return;
        }

        // Exclude the current parenthesis
        string excluded = s.substr(0, idx) + s.substr(idx + 1);
        solve(excluded, cnt - 1, idx, st);

        // Include the current parenthesis
        solve(s, cnt, idx + 1, st);
    }

    vector<string> removeInvalidParentheses(string s) {
        int open = 0, close = 0;
        for(char ch : s){
            if(ch == '(') open++;
            else if(ch == ')'){
                if(open > 0) open--;
                else close++;
            }
        }

        unordered_set<string> st;
        solve(s, open + close, 0, st);

        if(st.empty()) return {""};

        return vector<string>(st.begin(), st.end());
    }
};

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