<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>

from heapq import heapify, heappush, heappop
class Solution:
    def removeInvalidParentheses(self, s: str) -> List[str]:
        def is_valid(u):
            count = 0
            for char in u:
                if char == '(':
                    count += 1
                elif char == ')':
                    count -= 1
                # if close before open:
                if count < 0:
                    return False
            return count == 0
        min_edits = float('inf')
        queue = deque([s])
        visited = set() # s can have consecutive and repeated chars that lead to same string when yanked
        output = []
        while queue:
            u = queue.popleft()
            if is_valid(u):
                n_edits = len(s) - len(u)
                if n_edits <= min_edits:
                    heappush(output, (n_edits, u))
                    min_edits = n_edits
                continue
            for i in range(len(u)):
                if u[i] in '()':
                    t = u[:i] + u[i+1:]
                    if t not in visited:
                        queue.append(t)
                        visited.add(t)
        final_output = []
        while output and output[0][0] == min_edits:
            final_output.append(output.pop()[1])
        return final_output
20400