<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]:

        def get_counts(s):
            l, r = 0, 0
            for i in s:
                if i == "(":
                    l += 1
                elif i == ")":
                    if l == 0:
                        r += 1
                    else:
                        l -= 1
            return l, r

        def valid(s):
            count = 0
            for i in s:
                if i == "(":
                    count += 1
                elif i == ")":
                    count -= 1
                if count < 0:
                    return False
            return True
        
        result = []

        def dfs(s, start, l, r):
            if l == 0 and r == 0 and valid(s):
                result.append(s)
                return
            
            for i in range(start, len(s)):
                if i > start and s[i] == s[i - 1]:
                    continue
                
                if r > 0 and s[i] == ")":
                    dfs(s[:i] + s[i + 1:], i, l, r - 1) # delete )
                elif l > 0 and s[i] == "(":
                    dfs(s[:i] + s[i + 1:], i, l - 1, r) # delete (
        
        l, r = get_counts(s)
        dfs(s, 0, l, r)

        return result
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