<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):
            count = 0
            for c in s:
                if c == '(':
                    count +=1
                elif c == ')':
                    count -=1
                if count < 0:
                    return False
            return count == 0
        

        bracketset = set()
        q = deque([s])
        ret = []
        prev = -float(inf)

        while q:
            size = len(q)
            for _ in range(len(q)):
                curstring = q.popleft()
                if isvalid(curstring):
                    if len(curstring) >= prev:
                        prev = len(curstring)
                        ret.append(curstring)
                    break
                
                for i in range(len(curstring)):
                    if curstring[i].isalpha(): continue
                    newstring = curstring[:i] + curstring[i+1:]
                    if newstring not in bracketset:
                        q.append(newstring)
                        bracketset.add(newstring)
        
        return ret

                
        
20500