<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 __init__(self):
        self.res = []
        self.track = []

    def removeInvalidParentheses(self, s: str) -> List[str]:
        self.backtrack(s, 0)
        maxLen = 0
        for par in self.res:
            maxLen = max(maxLen, len(par))
        longest = set()
        for par in self.res:
            if len(par) == maxLen:
                longest.add(par)
        return list(longest)
    
    def backtrack(self, s, i):
        if i == len(s):
            if self.isValid("".join(self.track)):
                self.res.append("".join(self.track))
            return
        c = s[i]
        if c != "(" and c != ")":
            self.track.append(c)
            self.backtrack(s, i+1)
            self.track.pop()
        else:
            # Situation 1: Parentheses added in the new round
            self.track.append(c)
            self.backtrack(s, i+1)
            self.track.pop()
            # Situation 2: Parentheses NOT added in the new round
            self.backtrack(s, i+1)
    
    def isValid(self, s):
        left = 0
        for ch in s:
            if ch == "(":
                left += 1
            elif ch == ")":
                left -= 1
                if left < 0:
                    return False
        return left == 0
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