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Stack practicing solution in Clear category for Brackets by Sebastian.M
def checkio(expression):
stack = []
counter = -1
for x in expression:
if(x=="(" or x=="[" or x=="{"):
stack.append(x)
counter+=1
if(x==")"):
if(len(stack)>0):
if(stack[counter]=="("):
del stack[counter]
counter-=1
else:
stack.append(x)
else:
return False
if(x=="]"):
if(len(stack)>0):
if(stack[counter]=="["):
del stack[counter]
counter-=1
else:
stack.append(x)
else:
return False
if(x=="}"):
if(len(stack)>0):
if(stack[counter]=="{"):
del stack[counter]
counter-=1
else:
stack.append(x)
else:
return False
if(counter!=-1 or len(stack)!=0):
return False
else:
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. 12, 2016
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