<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 isvalid(s):
            cnt=0
            for i in s:
                if i=='(':
                    cnt+=1
                elif i==')':
                    if cnt>0:
                        cnt-=1
                    else:
                        return False
                
            return cnt==0

        def backtracking(s,ind,left,right):
            if left==right==0:
                if isvalid(s):
                    res.add(s)
                return

            for i in range(ind,len(s)):
                if s[i]=='(' and left>0:
                    backtracking(s[:i]+s[i+1:],i,left-1,right)

                if s[i]==')' and right>0:
                    backtracking(s[:i]+s[i+1:],i,left,right-1)

        left=right=0
        res=set()
        for i in s:
            if i=='(':
                left+=1
            elif i==')':
                if left>0:
                    left-=1
                else:
                    right+=1
        
        backtracking(s,0,left,right)
        return list(res)

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