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Classic and concise stack based solution solution in Clear category for Brackets by lrnx
def checkio(expression):
pairs = {'}':'{',']':'[',')':'('}
stack = []
for i in expression:
if i in pairs.values():
stack.append(i)
elif i in pairs.keys() and pairs[i] != (stack.pop() if stack else None):
return False
return stack == []
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
#print(checkio("(((1+(1+1))))]"))
assert checkio("((5+3)*2+1)") == True, "Simple"
assert checkio("{[(3+1)+2]+}") == True, "Different types"
assert checkio("(3+{1-1)}") == False, ") is alone inside {}"
assert checkio("[1+1]+(2*2)-{3/3}") == True, "Different operators"
assert checkio("(({[(((1)-2)+3)-3]/3}-3)") == False, "One is redundant"
assert checkio("2+3") == True, "No brackets, no problem"
Dec. 26, 2020
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