<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 is_valid(s):
        #     left_b = 0
        #     for i in s:
        #         if i == '(':
        #             left_b += 1
        #         elif i == ')':
        #             left_b -= 1
        #         if left_b < 0:
        #             return False

        #     return left_b == 0
        
        # res_dict = defaultdict(set)

        # def dfs(i, remove, curr):
        #     if i == len(s):
        #         if is_valid(curr):
        #             res_dict[remove].add(curr)
        #         return
            
        #     if s[i] in "()":
        #         dfs(i + 1, remove, curr + s[i])
        #         dfs(i + 1, remove + 1, curr)
        #     else:
        #         dfs(i + 1, remove, curr + s[i])

        # dfs(0, 0, '')
        # min_remove = min(res_dict.keys())
        # return list(res_dict[min_remove])
        def is_valid(s):
            left_b = 0
            for i in s:
                if i == '(':
                    left_b += 1
                elif i == ')':
                    left_b -= 1
                if left_b < 0:
                    return False

            return left_b == 0
        
        
        q = deque([s])
        visited = set()
        res = []
        traverse = True
        while q:
            candidate = q.popleft()
            if candidate not in visited:
                visited.add(candidate)
                
                if is_valid(candidate):
                    res.append(candidate)
                    traverse = False
                
                if traverse:
                    for i, p in enumerate(candidate):
                        if p in "()":
                            q.append(candidate[:i] + candidate[i + 1:])

        return res
    


           

            

        
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