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First solution in Clear category for Brackets by szymonkukla1
def checkio(expression):
i = 0
stos = []
while(i < len(expression)):
x = ord(expression[i])
if(x == 40):
stos.append(41)
if(x == 91):
stos.append(93)
if(x == 123):
stos.append(125)
if(x == 41 or x == 93 or x == 125):
if(len(stos) == 0):
return False
if(stos.pop() - x != 0):
return False
i += 1
if(len(stos) != 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. 16, 2016
