<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]:
        res = set()
        l, r, mr = 0, 0, 0
        for char in s:
            if char == "(":
                l += 1
            elif char == ")":
                l -= 1
            if l < 0:
                mr -= l
                l = 0
        mr += l 

        
        n = len(s)

        @lru_cache(None)
        def helper(i, curr, l, count):
            if l < 0:
                return
            if count > mr:
                return
            if i == n:
                if l == 0:
                    res.add(curr[:])
                return

            if s[i] == '(':
                helper(i+1, curr+s[i], l+1, count)
                helper(i+1, curr, l, count+1)
            elif s[i] == ')':
                helper(i+1, curr, l, count+1)
                helper(i+1, curr+s[i], l-1, count)
            else:
                helper(i+1, curr+s[i], l, count)

        helper(0, '', 0, 0)
        return list(res)
        
18900