[
stack
brackets
]
BinarySearch 0002 Balanced Brackets
Problem statement
https://binarysearch.com/problems/Balanced-Brackets/
Solution
Classical stack problem, easy version of 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