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More redundancy solution in Clear category for Brackets by Adrian_Mizielinski
def checkio(expression):
brackets = []
for ch in expression:
if ch in "([{":
brackets.append(ch)
elif ch==")":
if len(brackets)!=0:
if brackets.pop()!="(":
return False
else:
return False
elif ch=="]":
if len(brackets)!=0:
if brackets.pop()!="[":
return False
else:
return False
elif ch=="}":
if len(brackets)!=0:
if brackets.pop()!="{":
return False
else:
return False
if len(brackets)!=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. 4, 2016
