<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(s: str) -> bool:
            cnt = 0
            for c in s:
                if c == '(':
                    cnt += 1
                elif c == ')':
                    cnt -= 1
                if cnt < 0:
                    return False
            return cnt == 0
        res = []
        q = [s]
        checked = set()
        while q:
            n = len(q)
            foundValid = False
            for i in range(n):
                ss = q.pop(0)
                if ss in checked:
                    continue
                checked.add(ss)
                if isValid(ss):
                    foundValid = True
                    res.append(ss)
                    continue
                for i in range(len(ss)):
                    if ss[i] in '()':
                        ts = ss[:i] + ss[i+1:]
                        q.append(ts)
            if foundValid:
                break
        return res
                

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