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First solution in Clear category for Brackets by KarolinaK
def pair(a,b):
if((a=="(" and b==")") or (a=="[" and b=="]") or (a=="{" and b=="}")):
return True
else:
return False
def checkio(expression):
st = []
for i in expression:
if(i is "(" or i is "[" or i is "{"):
st.append(i)
print(st[len(st)-1])
if( i is ")" or i is "]" or i is "}"):
if(len(st)==0):
return False
elif(pair(st[len(st)-1],i) is False):
return False
else:
del st[-1]
print('\n')
if(len(st)==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. 19, 2016
