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First solution in Clear category for Brackets by Kamil0320
def checkio(expression):
br=["(","[","{","}","]",")"]
exp=list(expression)
br_exp = []
for i in exp:
if i in br:
br_exp.append(i)
if (pairer(br_exp, br) and next_char(br_exp, br)):
return True
return False
def pairer(list, br):
for q in range (3):
if ((list.count(br[q])+list.count(br[5-q]))%2)!=0:
return False
return True
def next_char(list, br):
for e in range (len(list)-1):
ind = 5 - br.index(list[e])
if list[e] in br[:3]:
if list[e+1] in (br[3:ind] or br[ind+1:6]):
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"
Oct. 28, 2016