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First solution in Speedy category for Brackets by Ch0bits
def checkio(expression):
pairs = {')': '(', ']': '[', '}': '{'}
stack = []
for x in expression:
if x in pairs.values():
stack.append(x)
elif x in pairs.keys():
if not stack:
return False # found excess closing brackets
# closing bracket must be the same kind as the opening bracket
if stack.pop() != pairs[x]:
return False
if stack:
return False # found too much brackets
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"
March 13, 2014
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