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First solution in Clear category for Brackets by DominikKossinski
def checkio(expression):
expression = list(expression)
nawiasy=[]
for i in expression:
if(i=="(" or
i=="[" or
i=="{"):
nawiasy.append(i)
if i==")":
if(len(nawiasy)==0 or nawiasy[-1]!="("):
return False
else:
nawiasy.pop()
if i=="]":
if(len(nawiasy)==0 or nawiasy[-1]!="["):
return False
else:
nawiasy.pop()
if i=="}":
if(len(nawiasy)==0 or nawiasy[-1]!="{"):
return False
else:
nawiasy.pop()
n=len(nawiasy)
if(n==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"
Oct. 24, 2016
