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Balanced Parentheses Check problem
By Subham, 2 years ago
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1
Given a string of opening and closing parentheses, check whether it’s balanced. We have 3 types of parentheses: round brackets: (), square brackets: [], and curly brackets: {}. Assume that the string doesn’t contain any other character than these, no spaces words or numbers. As a reminder, balanced parentheses require every opening parenthesis to be closed in the reverse order opened. For example ‘([])’ is balanced but ‘([)]’ is not.
You can assume the input string has no spaces.
Fill out your solution below:
def balance_check(s):
chars = []
matches = {')':'(',']':'[','}':'{'}
for c in s:
if c in matches:
if chars.pop() != matches[c]:
return False
else:
chars.append(c)
return chars == []
pass
balance_check('[]')
balance_check('[](){([[[]]])}')
balance_check('()(){]}')
Balanced parentheses
1 Answer
1
Neha
Solution :
First we will scan the string from left to right, and every time we see an opening parenthesis we push it to a stack, because we want the last opening parenthesis to be closed first. (Remember the FILO structure of a stack!)
Then, when we see a closing parenthesis we check whether the last opened one is the corresponding closing match, by popping an element from the stack. If it’s a valid match, then we proceed forward, if not return false.
Or if the stack is empty we also return false, because there’s no opening parenthesis associated with this closing one. In the end, we also check whether the stack is empty. If so, we return true, otherwise return false because there were some opened parenthesis that were not closed.
def balance_check(s):
# Check is even number of brackets
if len(s)%2 != 0:
return False
# Set of opening brackets
opening = set('([{')
# Matching Pairs
matches = set([ ('(',')'), ('[',']'), ('{','}') ])
# Use a list as a "Stack"
stack = []
# Check every parenthesis in string
for paren in s:
# If its an opening, append it to list
if paren in opening:
stack.append(paren)
else:
# Check that there are parentheses in Stack
if len(stack) == 0:
return False
# Check the last open parenthesis
last_open = stack.pop()
# Check if it has a closing match
if (last_open,paren) not in matches:
return False
return len(stack) == 0
balance_check('[]')
True
balance_check('[](){([[[]]])}')
True
balance_check('()(){]}')
False
