Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
Closing brackets stack solution in Speedy category for Brackets by romeech
# “Great!” Exclaimed Sofia. “Now we have the password.”
#
# “To what exactly?” Quipped Nikola.
#
# “Untold treasures, vast riches beyond belief! Gold! Silver! Silicon! Hydraulic Fluid!
# Anything your heart desires!”
#
# “And you’re going to do what with a password to absolutely nothing?” Nikola smirked.
#
# “Oh... Right...”
#
# Stephen spoke up. “Well, this door back here has a keypad.
# Only thing is the brackets look pretty busted up.
# We could try fixing it and then punching in the password?”
#
# “YES! That!” Sofia exclaimed.
#
# You are given an expression with numbers, brackets and operators.
# For this task only the brackets matter. Brackets come in three flavors: "{}" "()" or "[]".
# Brackets are used to determine scope or to restrict some expression.
# If a bracket is open, then it must be closed with a closing bracket of the same type.
# The scope of a bracket must not intersected by another bracket.
# In this task you should make a decision, whether to correct an expression
# or not based on the brackets.
# Do not worry about operators and operands.
#
# Input:An expression with different of types brackets as a string (unicode).
#
# Output:A verdict on the correctness of the expression in boolean (True or False).
#
# Precondition:
# There are only brackets ("{}" "()" or "[]"), digits or operators ("+" "-" "*" "/").
# 0 < len(expression) < 103
def checkio(expression):
closing_stack = []
for char in expression:
if char == '(':
closing_stack.append(')')
elif char == '[':
closing_stack.append(']')
elif char == '{':
closing_stack.append('}')
elif char in (')', ']', '}'):
if not closing_stack:
return False
closing_bracket = closing_stack.pop()
if char != closing_bracket:
return False
return len(closing_stack) == 0
# These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
assert checkio("((5+3)*2+1)") is True, "Simple"
assert checkio("{[(3+1)+2]+}") is True, "Different types"
assert checkio("(3+{1-1)}") is False, ") is alone inside {}"
assert checkio("[1+1]+(2*2)-{3/3}") is True, "Different operators"
assert checkio("(({[(((1)-2)+3)-3]/3}-3)") is False, "One is redundant"
assert checkio("2+3") is True, "No brackets, no problem"
Jan. 6, 2020
