<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.ans = set()

    def dfs(self, i, invalid_l, invalid_r, l_cnt, r_cnt, ans, s):
        if i == len(s):
            if invalid_l ==0 and invalid_r == 0:
                self.ans.add(ans)
            return
        if l_cnt < r_cnt:
            return
        if s[i] == "(" and invalid_l:
            self.dfs(i + 1, invalid_l - 1, invalid_r, l_cnt, r_cnt, ans, s)
        elif s[i] == ")" and invalid_r:
            self.dfs(i + 1, invalid_l, invalid_r - 1, l_cnt, r_cnt, ans, s) 
        self.dfs(
            i + 1,
            invalid_l,
            invalid_r,
            l_cnt + (1 if s[i] == "(" else 0),
            r_cnt + (1 if s[i] == ")" else 0),
            ans + s[i],
            s,
        )
        return

    def removeInvalidParentheses(self, s: str) -> List[str]:
        l, r = 0, 0
        for c in s:
            if c == "(":
                l += 1
            elif c == ")":
                if l:
                    l -= 1
                else:
                    r += 1
        self.dfs(0, l, r, 0, 0, "", s)
        return list(self.ans)

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