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using numbers solution in Clear category for Brackets by makosharkdb
def checkio(expression):
parenthesis = {'(':1, '{':2, '[':3, ')' : -1, '}':-2, ']':-3}
temp1 = [parenthesis.get(i) for i in expression if i in parenthesis.keys()]
prev = len(temp1)
post = 0
while prev != post:
prev = len(temp1)
for i in range(len(temp1)-1):
if temp1[i] > 0 and temp1[i+1] < 0:
if temp1[i] + temp1[i+1] == 0:
temp1[i:(i+2)] = []
break
post = len(temp1)
return len(temp1) == 0
#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"
Dec. 6, 2017
