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First solution in Uncategorized category for Brackets by krzysztof.gonda
def checkio(expression):
r = 0
b = 0
c = 0
w8 = []
for x in range(0,len(expression)):
if(expression[x]=='('):
w8.append(')')
r+=1
if(expression[x]=='{'):
w8.append('}')
c+=1
if(expression[x]=='['):
w8.append(']')
b+=1
if(expression[x]==')'):
if(r==0):
return False
else:
if(w8.pop() == ')'):
r-=1
else:
return False
if(expression[x]=='}'):
if(c==0):
return False
else:
if(w8.pop() == '}'):
c-=1
else:
return False
if(expression[x]==']'):
if(b==0):
return False
else:
if(w8.pop() == ']'):
b-=1
else:
return False
if(c==b==r==0):
return True
return False
#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. 22, 2017
