<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 is_valid_parencies(self, s: str) -> bool:
        count = 0

        for val in s:
            if val == "(":
                count += 1
            if val == ")":
                count -= 1
            if count < 0:
                return False

        return count == 0

    def removeInvalidParentheses(self, s: str) -> List[str]:
        def get_variation(idx: int, variation: List[str], o: int, c: int) -> List[str]:
            if c > o:
                return []

            if idx == len(s):
                return ["".join(variation)] if c == o else []

            variations = []

            variation.append(s[idx])
            if s[idx] == "(":
                variations = get_variation(idx + 1, variation, o + 1, c)
            elif s[idx] == ")":
                variations = get_variation(idx + 1, variation, o, c + 1)
            else:
                variations = get_variation(idx + 1, variation, o, c)

            variation.pop()

            if s[idx] == "(" or s[idx] == ")":
                variations += get_variation(idx + 1, variation, o, c)

            return variations

        all_variations = get_variation(0, [], 0, 0)

        # valid_variations = []

        # for variation in all_variations:
        #     if self.is_valid_parencies(variation):
        #         valid_variations.append(variation)

        all_variations.sort(key=lambda s: len(s), reverse=True)

        # print(f"{valid_variations=}")

        biggest_valid_combo = len(all_variations[0])
        result = set()

        for combo in all_variations:
            if len(combo) == biggest_valid_combo:
                result.add(combo)
            else:
                return result

        return result

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