Finding unique elements from the list of given numbers

Question

I have written a code which finds unique elements from the list of integers.

def Distinct(arr, n):

    for i in range(0, n):

        d = 0
        for j in range(0, i):
            if (arr[i] == arr[j]):
                d = 1
                break

        if (d == 0):
            print(arr[i], end=',')
    

n = int(input('Enter length of numbers: '))
arr = []
for i in range(n):
    a = input('enter the number: ')
    arr.append(a)
print(Distinct(arr, n)) 

if my given input array is [1,2,2,3,4] where n = 5 i get the output as 1,2,3,4,None but i want the output to be 1,2,3,4 can anyone tell how to fix this?

Answer 1

Your function Distinct has no return statement so it will always return None . When you call print(Distinct(arr, n)) this is printing the None that is returned. You can simply remove the print() like so:

Distinct(arr, n)

To remove the last comma you can't print a comma on the first iteration, then print the commas before the items. This is because you don't know which item will be the last one.

if d == 0 and i == 0:
    print(arr[i], end='')
elif d == 0:
    print("," + arr[i], end='')

Answer 2

Try something like this:

def Distinct(arr, n):
    lst = []
    for i in range(0, n):
        for j in range(0, i):
            if (arr[i] == arr[j]):
                lst.append(arr[i])
                break
    print(*lst, sep=',')
    

n = int(input('Enter length of numbers: '))
arr = []
for i in range(n):
    a = input('enter the number: ')
    arr.append(a)
Distinct(arr, n)

So first you compute all the numbers, and only at the end you print them