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Defaultdict are gems solution in Clear category for Brackets by furtadobb
from collections import defaultdict
keys = {')': '(', ']': '[', '}': '{'}
def checkio(expression):
pos = list()
d = defaultdict(list)
for i, l in enumerate(expression):
if l == '(' or l == '[' or l == '{':
d[l].append(i)
pos.append(i)
if l in keys:
if len(d[keys[l]]) > 0:
d[l].append(i)
pos.append(i)
else:
return False
out = ''.join(expression[i] for i in range(len(expression)) if i in pos)
if any([i for i in ['(]', '(}', '[)', '[}', '{)', '{]'] if i in out]) or len(out) % 2 != 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"
Dec. 21, 2020
