<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]:
        res = set()
        
        def check_valid(s):
            stack = deque()
            for i in range(len(s)):
                ch = s[i]
                if ch == '(':
                    stack.append(ch)
                elif ch == ')':
                    if stack and stack[-1] == '(':
                        stack.pop()
                    else:
                        stack.append(')')
            return len(stack) == 0, stack
                    
        stack = check_valid(s)[1]
        left = sum([1 for ch in stack if ch == '('])
        right = len(stack) - left
        
        res = set()
        found = set()
        
        def dfs(s, l, r):
            found.add(s)
            if l == r == 0 and check_valid(s)[0]: res.add(s)
                
            for i, ch in enumerate(s):
                if ch != '(' and ch != ')': continue
                if ch == '(' and l == 0: continue
                if ch == ')' and r == 0: continue
                
                rm = s[:i] + s[i + 1:]
                if rm not in found:
                    dfs(rm, l - (ch == '('), r - (ch == ')'))
                
        dfs(s, left, right)
        return list(res)
        
18700