<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]:
        
        @cache
        def isValid(s):
            stack = []
            for c in s:
                if c == "(":
                    stack.append(c)
                elif c == ")":
                    if stack and stack[-1] == "(":
                        stack.pop()
                    else:
                        stack.append(c)
            return stack

        res = []
        @cache
        def dfs(s, left, right):
            if left == 0 and right == 0:
                if isValid(s) == []:
                    res.append(s[:])
                    return
            else:
                for i, c in enumerate(s):
                    if c != "(" and c != ")":
                        continue

                    if (c == "(" and left == 0) or (c == ")" and right == 0):
                        continue

                    dfs(s[:i]+s[i+1:], left-(c=="("), right-(c==")"))
            return

        invalid = isValid(s)
        left = sum([1 for c in invalid if c == "("])
        right = len(invalid) - left
        dfs(s, left, right)
        return res
18100