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First solution in Clear category for Brackets by voxes96
def checkio(exp):
arr = []
op = ["(", "[", "{"]
cl = [")", "]", "}"]
for i in range(len(exp)):
if(exp[i] in op):
arr.append(exp[i])
if(exp[i] in cl):
if(len(arr) == 0):
return False
#if(exp[i] != arr.pop()):
if((arr[-1]=="(" and exp[i]!=")") or (arr[-1]=="[" and exp[i]!="]") or (arr[-1]=="{" and exp[i]!="}")):
return False
arr.pop()
if(len(arr) == 0):
return True
else:
return False
'''
for i in range(len(exp)):
if(exp[i] in ["(", "[", "{", "}", "]", ")"]):
arr.append(exp[i])
if(len(arr)%2 == 1):
return False
l = len(arr)
for i in range((l//2)+1):
if((arr[i]=="(" and arr[-(i+1)]!=")") or (arr[i]=="[" and arr[-(i+1)]!="]") or (arr[i]=="{" and arr[-(i+1)]!="}")):
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"
Oct. 29, 2016
