<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 search(string& s, int i, int numLeft, int numRight, int numInvalidLeft, int numInvalidRight, unordered_set<string>& answers, const string& prefix) {
        if (i >= s.size()) {
            if (numInvalidLeft == 0 && numInvalidRight == 0) {
                answers.insert(prefix);
            }
            return;
        }

        auto left = s[i] == '(';
        auto right = s[i] == ')';
        if (!left && !right) {
            search(s, i + 1, numLeft, numRight, numInvalidLeft, numInvalidRight, answers, prefix + s[i]);
        } else if (left) {
            if (numInvalidLeft > 0) {
                // try skipping this one
                search(s, i + 1, numLeft, numRight, numInvalidLeft-1, numInvalidRight, answers, prefix);
            }
            search(s, i + 1, numLeft+1, numRight, numInvalidLeft, numInvalidRight, answers, prefix + s[i]);
        } else {
            if (numInvalidRight > 0) {
                // try skipping this one
                search(s, i + 1, numLeft, numRight, numInvalidLeft, numInvalidRight-1, answers, prefix);
            }
            if (numLeft > numRight) {
                search(s, i + 1, numLeft, numRight+1, numInvalidLeft, numInvalidRight, answers, prefix + s[i]);
            }
        }
    }
    
    vector<string> removeInvalidParentheses(string s) {
        // figure out number of parens to remove, right and left
        // maintain a set of answers
        // recursive traversal
        //   base case: at end of the string: found a result if have eliminated extra parens. register it
        //   if char is not paren, push it on and continue to next elem
        //   otherwise char is a paren
        //      if it's a left paren
        //          if we have any invalid left parens left, try recursing to next char with the paren dropped
        //          also try recursing to next char including next paren
        //      if it's a right paren
        //          if we have any invalid right parens left AND num left parens > current num right parens: try recursing with this paren removed
        //          try recursing with this paren included

        int numLeftParens = 0, numRightParens = 0;
        for (auto c : s) {
            if (c == '(') {
                numLeftParens++;
            } else if (c == ')') {
                if (numLeftParens > 0) numLeftParens--;
                else numRightParens++;
            }
        }

        unordered_set<string> answers;
        search(s, 0, 0, 0, numLeftParens, numRightParens, answers, "");

        vector<string> result(answers.begin(), answers.end());
        if (result.empty()) {
            result.push_back("");
        }
        return result;
    }
};
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