<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:
    void DFS(int ind, int l, int r, const string &s, vector<string> &ret, string &path) {
    // l is the number of '('s and r is the number of ')'s in path
	if(r>l) return;
	
	if(ind==s.size()) {
	    // if !ret.empty(), we find one longest valid string
		if(l==r && (ret.empty() || path.size()==ret[0].size())) 
		    ret.push_back(path);
		return;
	}
	
	if(!ret.empty())
	    if(path.size() + s.size() - ind < ret[0].size()) return;
	
	char c = s[ind++];
	path.push_back(c);
	if(c!='(' && c!=')')
		DFS(ind,l,r,s,ret,path);
	else if(c=='(')
		DFS(ind,l+1,r,s,ret,path);
	else 
		DFS(ind,l,r+1,s,ret,path);
	
	path.pop_back();
	if(c!='(' && c!=')')
	    DFS(ind,l,r,s,ret,path);
	else {
	    //once decide to give up current '(' or ')', we do not need to try the continuous '('s or ')'s
	    while(c==s[ind]) ++ind;
	    DFS(ind,l,r,s,ret,path);
	}
}

vector<string> removeInvalidParentheses(string s) {
	vector<string> ret;
	string path;
	DFS(0,0,0,s,ret,path);
	if(ret.empty()) return {""};
	return ret;
}
};
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