def extractlyEvenlyDivisibleHelp(inputlst, number, outputlst = []):
        if number == 0:
            print([])
            return
        if inputlst[0] % number == 0:
            outputlst.append(inputlst[0])
            outputlst.sort()
    
    
        inputlst.remove(inputlst[0])
    
        if len(inputlst) == 0:
            print(outputlst)
        else:
            extractlyEvenlyDivisibleHelp(inputlst, number, outputlst)
    
    
    def extractlyEvenlyDivisible(input, number):
        extractlyEvenlyDivisibleHelp(input, number, [])

extractlyEvenlyDivisible([1,2,3,4,5,6,7,8,9], 0)

extractlyEvenlyDivisible([1,2,-3,4,5,-6,7,8,9,9,6], -3)

extractlyEvenlyDivisible([1,2,3,4,5,6,7,8,9,10, 10], 5)

Output:

[ ]

[-6, -3, 6, 9, 9]

[5, 10, 10]

Expected output:

[ ]

[-6, -3, 6, 9]

[5, 10]

I need help, as I need only one 9, I mean if I input couple times the same digit, it should return only one time

CodePudding user response：

You could filter out duplicates from the input list before passing to the helper using set:

def extractlyEvenlyDivisible(inputlst, number):
    extractlyEvenlyDivisibleHelp(list(set(inputlst)), number, [])

Also it is bad practice to redefine built-ins like input even in a function, I would change this to a different variable name