<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 collections import deque

class Solution:
    def isValid(self, s: str) -> bool:
        count = 0
        for char in s:
            if char == '(':
                count += 1
            elif char == ')':
                count -= 1
            if count < 0:
                return False
        return count == 0

    def removeInvalidParentheses(self, s: str) -> list:
        if not s:
            return [""]
        
        # Initialize BFS
        result = []
        queue = deque([s])
        visited = set([s])
        found = False
        
        while queue:
            current = queue.popleft()
            
            # Check if the current string is valid
            if self.isValid(current):
                result.append(current)
                found = True  # We've found valid strings, so we don't need to remove more parentheses
            elif not found:
                # Generate all possible strings by removing one parenthesis at a time
                for i in range(len(current)):
                    if current[i] not in ('(', ')'):
                        continue
                    next_string = current[:i] + current[i+1:]
                    if next_string not in visited:
                        visited.add(next_string)
                        queue.append(next_string)
        
        return result

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