<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]:
        if not s:
            return [""]

        res = []
        stack = [s]
        seen = set()

        while stack:
            new_stack = []

            for possible in stack:
                if self.isValid(possible):
                    res.append(possible)

                else:
                    for j in range(len(possible)):
                        new_word = possible[:j] + possible[j+1:]
                        if new_word not in seen:
                            seen.add(new_word)
                            new_stack.append(new_word)

            if res:
                return res

            stack = new_stack
            
    def isValid(self, s):
        stack = []
        for ch in s:
            if ch == "(":
                stack.append(0)

            elif ch == ")":
                if stack:
                    stack.pop()
                else:
                    return False
        
        return stack == []
20200