<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]:
        left = 0
        total = 0
        for i in s:
            if i == '(':
                left += 1
                total += 1
            elif i == ')' and left > 0:
                left -= 1
        target = total - left
        def helper(i, left, right, target, curr):
            if i == len(s):
                if left == right == target:
                    # print(left, right, target, curr)
                    res.add(''.join(curr))
                return
            
            if s[i] not in '()':
                curr.append(s[i])
                helper(i + 1, left, right, target, curr)
                curr.pop()
            elif s[i] == '(':
                # keep
                if left < target:
                    curr.append(s[i])
                    helper(i + 1, left + 1, right, target, curr)
                    curr.pop()
                # not keep
                helper(i + 1, left, right, target, curr)
            else:
                # keep
                if right < left:
                    curr.append(s[i])
                    helper(i + 1, left, right + 1, target, curr)
                    curr.pop()
                # not keep
                helper(i + 1, left, right, target, curr)

        res = set()
        helper(0, 0, 0, target, [])
        return res
16800