Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
First solution in Clear category for Brackets by karolk10
def checkio(expression):
tab = []
for i in range(len(expression)):
if expression[i] == "(":
tab = tab + ["("]
if expression[i] == "[":
tab = tab + ["["]
if expression[i] == "{":
tab = tab + ["{"]
if expression[i] == ")":
if len(tab) == 0:
return False
else:
if tab[len(tab)-1] == "(":
tab = tab[:len(tab)-1]
else:
return False
if expression[i] == "]":
if len(tab) == 0:
return False
else:
if tab[len(tab)-1] == "[":
tab = tab[:len(tab)-1]
else:
return False
if expression[i] == "}":
if len(tab) == 0:
return False
else:
if tab[len(tab)-1] == "{":
tab = tab[:len(tab)-1]
else:
return False
if len(tab) == 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. 15, 2016
