<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 min_edits(s) -> int:
            closing = 0
            opening = 0
            for char in s:
                if char == "(":
                    opening += 1
                elif char == ")" and opening == 0:
                    closing += 1
                elif char == ")":
                    opening -= 1
            return closing + opening
       
        def valid_paren(s) -> bool:
            opening = 0
            for char in s:
                if char == "(":
                    opening += 1
                elif char == ")" and opening == 0:
                    return False
                elif char == ")":
                    opening -= 1
            return opening == 0
       
        total_edits = min_edits(s)
       
        visited = set()
        queue = deque()
        queue.append((s, total_edits))
       
        res = []
       
        while queue:
            st, edits = queue.popleft()
           
            if st in visited:
                continue
            visited.add(st)
           
            if edits == 0 and valid_paren(st):
                res.append(st)
            elif edits == 0:
                continue
           
            for i in range(len(st)):
                if st[i] not in "()":
                    continue
                new_str = st[:i] + st[i + 1:]
                queue.append((new_str, edits - 1))
       
        return res
19800