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LC 1249 [M] Minimum Remove to Make Valid Parentheses
Code with Senpai edited this page Mar 13, 2023
·
21 revisions
intuitive? Y
(NOT a recursion/backtracking/DP problem but the classic stack problem!)
- invalid open indexes stack, invalid close indexes stack
- push opens
- on close, pop open if there is one, else push close
- combine the lone opens and closes indexes, and return the string without them
T: O(n)
S: O(n)
there is a S: O(1) solution if you just count but it's probably not needed
class Solution:
def minRemoveToMakeValid(self, s: str) -> str:
l, r = [], []
for i, x in enumerate(s):
if x == '(':
l.append(i)
elif x == ')':
if l:
l.pop() # valid, close a pair
else:
r.append(i) # invalid
# unpaired = set(l + r) # you don't really need a set but it speeds up time, list is ok
unpaired = l + r
return [x for i, x in enumerate(s) if i not in unpaired]
class Solution:
def minRemoveToMakeValid(self, s) :
stack=[]
split_str=list(s)
for i in range(len(s)):
if s[i]=='(':
# if current char is '(' then push it to stack
stack.append(i)
elif s[i]==')':
# if current char is ')' then pop top of the stack
if len(stack) !=0:
stack.pop()
else:
# if our stack is empty then we can't pop so make current char as ''
split_str[i]=""
for i in stack: # the invalid parens
split_str[i]=""
return '' .join(split_str)
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