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First solution in Clear category for Brackets by 9maksim
def checkio(expression):
from re import findall
a = findall (r"[\(\)\{\}\[\]]",expression)
d = {")":"(" , "]":"[" , "}":"{"}
for j in range(len(a)//2):
for i in range(1,len(a)):
if a[i] in d:
if d[a[i]] == a[i - 1]:
del a[i]
del a[i - 1]
break
return not a
#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. 23, 2019
