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First solution in Uncategorized category for Brackets by danikmil
# migrated from python 2.7
def checkio(expression):
stack = []
ex = [a for a in expression if a in ('(', ')', '{', '}', '[', ']')]
if not ex:
return True
if len(ex) % 2:
return False
for item in expression:
if item in ('(', '{', '['):
stack.append(item)
if not stack and item in (')', '}', ']'):
return False
if item == ')':
if stack[-1] == '(':
del stack[-1]
else:
return False
if item == '}':
if stack[-1] == '{':
del stack[-1]
else:
return False
if item == ']':
if stack[-1] == '[':
del stack[-1]
else:
return False
if not stack:
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"
Nov. 25, 2014
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