Given a string s that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return all the possible results. You may return the answer in any order.
Example 1:
Input: s = "()())()"
Output: ["(())()","()()()"]
Example 2:
Input: s = "(a)())()"
Output: ["(a())()","(a)()()"]
Example 3:
Input: s = ")("
Output: [""]
Constraints:
-
1 <= s.length <= 25 -
sconsists of lowercase English letters and parentheses'('and')'. - There will be at most
20parentheses ins.
SOLUTION:
class Solution:
def isValid(self, s: str) -> bool:
stack = []
valid = True
for bracket in s:
if bracket == '(':
stack.append(bracket)
elif bracket == ')':
if len(stack) > 0:
if stack.pop() + bracket != '()':
valid = False
break
else:
valid = False
break
if len(stack) > 0:
valid = False
return valid
def removeInvalidParentheses(self, s: str) -> List[str]:
n = len(s)
paths = [("", 0, 0)]
btoi = { '(': 1, ')': -1 }
i = 0
mindels = n
op = set()
visited = set()
while i < len(paths):
curr, j, currctr = paths[i]
if j == n:
if self.isValid(curr):
if n - len(curr) < mindels:
mindels = n - len(curr)
op = { curr }
elif n - len(curr) == mindels:
op.add(curr)
if j < n and j - len(curr) <= mindels:
nctr = currctr + btoi.get(s[j], 0)
if nctr >= 0 and (curr + s[j], j + 1) not in visited:
visited.add((curr + s[j], j + 1))
paths.append((curr + s[j], j + 1, nctr))
if s[j] == '(' or s[j] == ')' and (curr, j + 1) not in visited:
visited.add((curr, j + 1))
paths.append((curr, j + 1, currctr))
i += 1
return list(op)
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