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First solution in Clear category for Brackets by Krzysztof_bonczyk
def checkio(expression):
czy=False
listt=[]
print(len(listt))
for i in expression:
if i=='(':
listt.append(i)
elif i=='{':
listt.append(i)
elif i=='[':
listt.append(i)
if listt:
if (i==')')and(listt[len(listt)-1]=='('):
listt.pop()
elif (i=='}')and(listt[len(listt)-1]=='{'):
listt.pop()
elif (i==']')and(listt[len(listt)-1]=='['):
listt.pop()
if not listt:
czy=True
if expression.count('(')!=expression.count(')'):
czy=False
if expression.count('{')!=expression.count('}'):
czy=False
if expression.count('[')!=expression.count(']'):
czy=False
return czy
#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. 3, 2016
