<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 removeInvalidParentheses(self, s: str) -> List[str]:
        bfs = deque([s])
        seen, res = set(), []
        while bfs:
            # classic layer by layer bfs traversal
            for i in range(len(bfs)):
                curr_str = bfs.popleft()
                if self.isvalid(curr_str):
                    res.append(curr_str)
                if not res:
                    # at current level, we haven't found any valid curr_str yet
                    for j in range(len(curr_str)):
                        if curr_str[j] == '(' or curr_str[j] == ')':
                            next_str = curr_str[:j] + curr_str[j+1:]
                            if next_str not in seen:
                                seen.add(next_str)
                                bfs.append(next_str)
        return res
                
        
    def isvalid(self, s):
        dic = {'(': 1, ')': -1}
        count = 0
        for char in s:
            count += dic.get(char, 0)
            if count < 0:
                return False # early exit coz once count of ')' > '(', s can't be valid
        return count == 0
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