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First solution in Clear category for Brackets by Coamo
def checkio(expression):
otw = []
otw.append('#')
for x in expression:
if x in ['(', '[', '{']:
otw.append(x)
elif x == ')':
if otw[-1] == '(':
otw.pop()
else:
return False
elif x == ']':
if otw[-1] == '[':
otw.pop()
else:
return False
elif x == '}':
if otw[-1] == '{':
otw.pop()
else:
return False
if otw[-1] == '#':
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. 13, 2016
