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First solution in Clear category for Brackets by Elater
def checkio(expression):
a=list()
for i in expression:
if(i=="[" or i=="{" or i=="("): a.append(i)
elif((i=="]" or i=="}" or i==")") and a==[]): return False
elif (i=="]"):
if(a.pop()!="["): return False
elif (i=="}"):
if(a.pop()!="{"): return False
elif (i==")"):
if(a.pop()!="("): return False
if(a==[]): return True
return False
#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. 18, 2016
