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First solution in Clear category for Brackets by Wojtas1411
def checkio(expression):
x="0123456789+-*/"
exp=""
for i in range(len(x)):
for j in range(len(expression)):
if expression[j]!=x[i]:
exp+=expression[j]
expression=exp
exp=""
print(expression)
exp=""
pom=0
while expression.rfind('{}')!=-1 or expression.rfind('[]')!=-1 or expression.rfind('()')!=-1:
if expression.rfind('{}')!=-1:
pom = expression.rfind('{}')
for i in range(len(expression)):
if i!=pom and i!=pom+1:
exp+=expression[i]
expression=exp
exp=""
if expression.rfind('[]')!=-1:
pom = expression.rfind('[]')
for i in range(len(expression)):
if i!=pom and i!=pom+1:
exp+=expression[i]
expression=exp
exp=""
if expression.rfind('()')!=-1:
pom = expression.rfind('()')
for i in range(len(expression)):
if i!=pom and i!=pom+1:
exp+=expression[i]
expression=exp
exp=""
if len(expression)>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. 14, 2016
