<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:
    def __init__(self):
        self.valid_expressions = None
        self.min_removed = None
    
    def reset(self):
        self.valid_expressions = set()
        self.min_removed = float('inf')
    
    def backtrack(self, string, index, left, right, expr, remove_cnt):
        if index == len(string):
            if left == right:
                if remove_cnt == self.min_removed:
                    self.valid_expressions.add("".join(expr))
                elif remove_cnt < self.min_removed:
                    self.valid_expressions = set()
                    self.min_removed=remove_cnt
                    self.valid_expressions.add("".join(expr))
        else:
            cur = string[index]
            if cur != "(" and cur != ")":
                expr.append(cur)
                self.backtrack(string, index+1, left, right, expr, remove_cnt)
                expr.pop()
            else:
                self.backtrack(string, index+1, left, right, expr, remove_cnt+1)

                expr.append(cur)
                if cur == "(":
                    self.backtrack(string, index+1, left+1, right, expr, remove_cnt)
                elif left > right:
                    self.backtrack(string, index+1, left, right+1, expr, remove_cnt)
                expr.pop()

            
    
    def removeInvalidParentheses(self, s: str) -> List[str]:
        self.reset()
        self.backtrack(s, 0, 0, 0, [], 0)

        return list(self.valid_expressions)
        
16200