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# Runtime: 96 ms (Top 83.6%) | Memory: 18.67 MB (Top 9.9%)
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class Solution:
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    def removeInvalidParentheses(self, s: str) -> List[str]:
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        ## RC ##
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        ## APPROACH : BACK-TRACKING ##
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        ## Similar to Leetcode 32. Longest Valid Parentheses ##
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        ## LOGIC ##
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        #   1. use stack to find invalid left and right braces.
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        #   2. if its close brace at index i , you can remove it directly to make it valid and also you can also remove any of the close braces before that i.e in the range [0,i-1]
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        #   3. similarly for open brace, left over at index i, you can remove it or any other open brace after that i.e [i+1, end]
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        #   4. if left over braces are more than 1 say 2 close braces here, you need to make combinations of all 2 braces before that index and find valid parentheses.
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        #   5. so, we count left and right invalid braces and do backtracking removing them
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        ## TIME COMPLEXITY : O(2^N) ## (each brace has 2 options: exits or to be removed)
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        ## SPACE COMPLEXITY : O(N) ##
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        def isValid(s):
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            stack = []
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            for i in range(len(s)):
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                if s[i] == "(":
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                    stack.append((i, "("))
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                elif s[i] == ")":
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                    if stack and stack[-1][1] == "(":
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                        stack.pop()
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                    else:
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                        stack.append((i, ")"))  # pushing invalid close braces also
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            return len(stack) == 0, stack
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        def dfs(s, left, right):
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            visited.add(s)
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            if left == 0 and right == 0 and isValid(s)[0]:
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                res.append(s)
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            for i, ch in enumerate(s):
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                if ch != "(" and ch != ")":
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                    continue  # if it is any other char ignore.
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                if (ch == "(" and left == 0) or (ch == ")" and right == 0):
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                    continue  # if left == 0 then removing '(' will only cause imbalance. Hence, skip.
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                if s[:i] + s[i + 1 :] not in visited:
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                    dfs(s[:i] + s[i + 1 :], left - (ch == "("), right - (ch == ")"))
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        stack = isValid(s)[1]
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        lc = sum([1 for val in stack if val[1] == "("])  # num of left braces
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        rc = len(stack) - lc
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        res, visited = [], set()
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        dfs(s, lc, rc)
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        return res