Set list elements to 0 from an index to the end if partial sum is under certain value

Having a list of numbers, say:

numbers = [4, 5, 7, 2]

How could I write a function f that takes the list and a value and returns these results (I think the examples explains better):

f(numbers, 4)
[4, 0, 0, 0]

f(numbers, 3)
[0, 0, 0, 0]

f(numbers, 10)
[4, 5, 0, 0]

f(numbers, 17)
[4, 7, 7, 0]

I need to update the list with zeros from and index i such that sum(l[0:i]) <= val and sum(l[0:i+1]) > val.

I tried:

def f(numbers, val, last_index=-1):
    if sum(numbers[0:last_index]) > val:
         numbers[last_index] = 0
         last_index -= 1
         f(numbers, val, last_index)

I need to mutate the objects contained in the list. Not allowed to make copies. The function I’ve written does not work.

>Solution :

I would use a loop with enumerate rather than recursion:

def f(nums,target):
    total = 0
    for i,n in enumerate(nums):
        if total <= target: total += n
        if total > target:
            nums[i:] = [0]*(len(nums)-i)
            break

passes all of your test cases (after the typo is removed from the last).