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Clear and Easy <=== comment what do you think solution in Clear category for Brackets by piotr.sawicki
def checkio(data):
brackets = [['(', ')'],
['{', '}'],
['[', ']']]
stack = []
for i in data:
if (i in brackets[0][1] or i in brackets[1][1] or i in brackets[2][1]) and len(stack) == 0:
return False
if i in brackets[0][0] or i in brackets[1][0] or i in brackets[2][0]:
stack.append(i)
for k in range(len(brackets)):
if i in brackets[k][1] and stack[-1] == brackets[k][0]: stack.pop()
elif i in brackets[k][1] and stack[-1] != brackets[k][0]: return False
if len(stack) != 0:
return False
return True
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
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"
Nov. 17, 2017
