<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]:
        leftRem, rightRem = 0, 0
        for c in s:
            if c == '(':
                leftRem += 1
            elif c == ')':
                rightRem = rightRem + 1 if leftRem == 0 else rightRem
                leftRem = leftRem - 1 if leftRem > 0 else leftRem

        result = set()
        def dfs(i, leftCount, rightCount, leftRem, rightRem, cur):
            if i >= len(s):
                if leftRem == 0 and rightRem == 0:
                    answer = "".join(cur)
                    result.add(answer)
                return

            # Discard 
            if s[i] == '(' and leftRem > 0:
                dfs(i+1, leftCount, rightCount,  leftRem - 1, rightRem,  cur)
            elif s[i] == ')' and rightRem > 0:
                dfs(i+1, leftCount, rightCount,  leftRem, rightRem - 1,  cur)

            # Include
            cur.append(s[i])
            if s[i] != '(' and s[i] != ')':
                dfs(i+1, leftCount, rightCount, leftRem, rightRem, cur)
            elif s[i] == '(':
                dfs(i+1, leftCount+1, rightCount, leftRem, rightRem, cur)
            elif s[i] == ')' and leftCount > rightCount:
                dfs(i+1, leftCount, rightCount+1, leftRem, rightRem, cur)
            cur.pop()

        dfs(0, 0, 0, leftRem, rightRem, [])
        return list(result)

                


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