[
stack
]
BinarySearch 0036 Balanced Brackets Sequel
Problem statement
https://binarysearch.com/problems/Balanced-Brackets-Sequel/
Solution
Equal to Leetcode 20. Valid Parentheses
Complexity
It is O(n)
for time and space.
Code
class Solution:
def solve(self, s):
dct = {"(": ")", "{": "}", "[": "]"}
stack = []
for char in s:
if char in dct:
stack.append(char)
else:
if not stack or char != dct[stack.pop()]: return False
return not stack