<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 ['']
        
        left, right = 0, 0
        for i in range(len(s)): # calculate left, right costs O(N)
            if s[i] not in '()':
                continue
            if s[i]=='(':
                left+=1
            else:
                if left:
                    left-=1
                else:
                    right+=1
        
        def isValid(s):
            left, right = 0, 0
            for c in s:
                if c not in '()':
                    continue
                if c=='(':
                    left+=1
                else:
                    if left:
                        left-=1
                    else:
                        return False
            return left==0

        def helper(s, start,left,right):
            if left==0 and right==0 and isValid(s):
                res.append(str(s))
                return 
            for i in range(start, len(s)):
                # if i>start and s[i]==s[i-1]:
                #     continue
                if s[i] in '()':
                    ns=s[:i]+s[i+1:]
                    if right>0 and s[i]==')':
                        helper(ns, i, left, right-1)
                    if left>0 and s[i]=='(':
                        helper(ns, i, left-1, right)

            

            
        res=[]
        helper(s, 0, left, right)
        return list(set(res))
16800