<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) {
        if (r > l)
            return;

        if (ind == s.size()) {
            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 {
            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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