<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]:
        x=set()
        x.update(self.riph(s, [], 0, 0))
        y = sorted(x, key=lambda x: len(x), reverse=True)
        #print(len(y[0]))
        a = []
        for z in y:
            if len(z) == len(y[0]):
                a.append(z)
        return a


    def riph(self, s:str, pfx:list, num_open:int, num_close:int) -> List[str]:
        if not s:
            #print("num_close", num_close, "num_open", num_open)
            if num_open==0:
                return ["".join(pfx)]
            else:
                return []
        c = s[0]
        res = []
        prefix = [x for x in pfx]
        rem = []
        norem = []
        if c.isalpha():
            prefix.append(c)
            return self.riph(s[1:], prefix, num_open, num_close)
        else:
            if c == '(' or c == ')':
                if num_open > 0 or c=='(':
                    prefix.append(c)
                    norem = self.riph(s[1:], prefix, num_open+1 if c=="(" else num_open-1, num_close+1 if c=="(" else num_close-1)
                rem = self.riph(s[1:], pfx, num_open, num_close)
                
                #if len(rem)==len(norem):
                res.extend(rem)
                res.extend(norem)
                #elif len(rem) > len(norem):
                    #res.extend(rem)
                #else:
                    #res.extend(norem)
        return res



            
19000