Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
First solution in Clear category for Brackets by HubertDolny
def checkio(expression):
openingBrackets= ["(","[","{"]
closingBrackets=[")","]","}"]
table=[]
for i in range(len(expression)):
if expression[i] in openingBrackets:
table.append(openingBrackets[openingBrackets.index(expression[i])])
elif expression[i] in closingBrackets:
if(len(table)==0):
return False
elif not(table.pop()==openingBrackets[closingBrackets.index(expression[i])]):
return False
if not( len(table))==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"
Nov. 4, 2016
