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First solution in Clear category for Brackets by Luker
def checkio(expression):
brocets=[]
for i in range(0,len(expression)):
if expression[i]=='{':
brocets.append('{')
elif expression[i]=='(':
brocets.append('(')
elif expression[i]=='[':
brocets.append('[')
elif expression[i]=='}':
if len(brocets)==0:
return False
if brocets.pop()!='{':
return False
elif expression[i]==')':
if len(brocets)==0:
return False
if brocets.pop()!='(':
return False
elif expression[i]==']':
if len(brocets)==0:
return False
if brocets.pop()!='[':
return False
if len(brocets)!=0:
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"
Nov. 21, 2016
