<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 removeInvalidParentheses(self, s: str) -> List[str]:
        def isValid(parens):
            open, closed = 0, 0
            for i in parens:
                if i  == '(':
                    open += 1
                if i == ')':
                    closed += 1
                if closed > open:
                    return False
            return open == closed
        
        ans = []
        cur = [s]
        n = len(s)
        vis = set()
        while not ans:
            next = []
            for parens in cur:
                if isValid(parens):
                    ans.append(parens)
                else:
                    for i, ch in enumerate(parens):
                        if ch not in "()":
                            continue
                        tmp = parens[:i] + parens[i+1:]
                        if tmp in vis:
                            continue
                        next.append(tmp)
                        vis.add(tmp)
            cur = next


        return ans
17500