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First solution in Clear category for Brackets by ylrih
def checkio(expression):
bo = ["{}","()","[]"]
a = list(expression)
mask = a[:]
[mask.remove(a[i]) for i in range(len(a)) if a[i] not in "".join(bo)]
if len(mask) % 2 == 0:
mask = "".join(mask)
for i in range(len(mask)):
for j in bo:
mask = mask.replace(j, "")
if mask == "":
return True
else:
return False
else:
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"
July 1, 2016
