<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]:
        #需要计算有多少冗余括号
        #dfs去找最小n_to_delete的s
        self.res = []  
        self.visited = set([s])  #忘记了[]
        self.dfs(s, self.calc(s))
        return self.res  

    def dfs(self, s, n_to_delete):
        if n_to_delete == 0:
            return self.res.append(s)  
        
        for i in range(len(s)):
            if s[i] in ['(', ')']: #错写成== 结果res里面啥也没有
                new_s = s[:i] + s[i+1:]  
                new_n_to_delete = self.calc(new_s)  
                if new_s not in self.visited and new_n_to_delete < n_to_delete:
                    self.dfs(new_s, new_n_to_delete)
                    self.visited.add(new_s)
    
    def calc(self, s):
        need_l = need_r =0
        for c in s:
            if c == '(':
                need_r += 1
            elif c == ')':
                if need_r > 0:
                    need_r -= 1
                else:
                    need_l += 1
        return need_l + need_r
        
16500