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First solution in Clear category for Brackets by matesik1995
def checkio(expression):
que = []
for i in range(len(expression)):
if expression[i]=='{':
que.append('{')
elif expression[i]=='[':
que.append('[')
elif expression[i]=='(':
que.append('(')
if expression[i]=='}':
if len(que) > 0:
if que.pop()!='{':
return False
else:
return False
if expression[i]==')':
if len(que) > 0:
if que.pop()!='(':
return False
else:
return False
if expression[i]==']':
if len(que) > 0:
if que.pop()!='[':
return False
else:
return False
if len(que)!=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. 29, 2016
