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First solution in Clear category for Brackets by Michal_Cichy
def gole(expression):
zaba=""
nawiasy="({[]})"
for i in expression:
if i in nawiasy:
zaba=zaba+i
return zaba
def dodaj(stos,x):
stos.insert(0,x)
def zdejmij(stos):
if len(stos)==0:
return None
x=stos[0]
stos.remove(x)
return x
def domkniecie(a,b):
return (abs(ord(a)-ord(b))>2 or (ord(a)-ord(b))==0)
def checkio(e):
e=gole(e)
stos=[]
L="([{"
P=")]}"
for a in e:
if a in L:
dodaj(stos, a)
if a in P:
if len(stos)==0:
return False
if domkniecie(a, stos[0])==1:
return False
zdejmij(stos)
if len(stos)==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. 17, 2016
