unexpected result at python

Question

I tried to print a factor of a number using Python. ie if input value is 24 then the output is [1,2,2,2,3] (1*2*2*2*3=24).

But the answer of the below program is [1,2,3,2,2,2,2] instead of [1,2,3,2,2]

def is_prime(i):
    for j in range(2,i+1):
        if(i%j==0):
            break
    if(i==j):
        return 1
    else:
        return 0
def fact(a):
    i=2
    while(i<=a):
        if(a%i==0):
            if(is_prime(i)):
                li.append(i)
                a=int(a/i)
                if(a==1):
                    return 
            else:
                fact(a)
        i=i+1
    if(a>=2):
        fact(a)
    return
li=[]
li.append(1)
a=int(input())
fact(a)
print(li)

Answer 1

The problem is the recursive call:

while(i<=a):
    if(a%i==0):
        if(is_prime(i)):
            li.append(i)
            a=int(a/i)
            if(a==1):
                return 
        else:
            fact(a)  # <-- problem
    i=i+1

This will call fact again for any non-prime factor of a , like 4 , 6 , 8 or 12 in the case of 24 . And those calls will again add 2 to the (global) list of results. Instead, remove the recursive call and just continue with the loop. Also, you should change the outermost if to while so multiple of the same factor can be found.

Also, the call to is_prime is redundant, as i is already guaranteed to be prime (otherwise a would have already been divided by one of i 's prime factors), and the return is redundant, too. Finally, for this function to be applicable more than just once, I strongly suggest moving the declaration of li inside the function and returning it at the end.

def fact(a):
    li = [1]  # actually, 1 is not a prime factor, though
    i = 2
    while i <= a:
        while a % i == 0:
            li.append(i)
            a = a // i
        i += 1
    return li

li = fact(a)