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First solution in Clear category for Brackets by Lachesis_132296
#list.append(x) - add an item to the end of the list
#list.pop(i) - remove the item at the given position in the list, and return it.
#list.pop() - remove and return the last item in the list
def top(list):
i = len(list) -1
if i >= 0:
return list[i]
else:
return -1
def checkio(expression):
t = []
for b in expression:
if b == '{' or b == '(' or b == '[':
t.append(b)
if b == '}' or b == ')' or b == ']':
if (top(t), b) == ('{', '}') or (top(t), b) == ('(', ')') or (top(t), b) == ('[', ']'):
t.pop()
else:
return False
if len(t) == 0:
return True
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"
Nov. 19, 2016
