HTMLify
expression-contains-redundant-bracket-or-not.py
Views: 2 | Author: prakhardoneria
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | class Solution(): def checkRedundancy(self, s): st = [] for char in s: if char == ')': top = st.pop() is_redundant = True while top != '(': if top in '+-*/': is_redundant = False top = st.pop() if is_redundant: return True else: st.append(char) return False |