<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):
            stack=[]
            for c in s:
                if c=="(":
                    stack.append(c)
                elif c==")":
                    if not stack:
                        return False
                    else:
                        stack.pop()
            return len(stack)==0
        
        res=[]
        q=deque([s])
        seen=set([s])
        while q:
            for i in range(len(q)):
                next_string=q.popleft()
                if isValid(next_string):
                    res.append(next_string)
                for j in range(len(next_string)):
                    if next_string[j] in ("(",")"):
                        new_string=next_string[:j]+next_string[j+1:]
                        if new_string in seen:
                            continue
                        q.append(new_string)
                        seen.add(new_string)
            if res:
                return res

        
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