My attempt to extend the “divide product by array element” approach to arrays having one or more zeros:

```
def findProduct(arr):
has_one_zero = False
prod = 1
for n in arr:
if n != 0:
prod *= n
else:
if has_one_zero:
return [0] * len(arr) # more than one zero
has_one_zero = True
zero_idx = arr.index(n)
continue
if has_one_zero:
l_out = [0] * len(arr)
l_out[zero_idx] = prod
return l_out
else:
return [prod//arr[idx] for idx in range(len(arr))]
print(findProduct([1,2,3,6, 4]))
print(findProduct([1,2,3,0, 4]))
print(findProduct([1,0,3,0, 4]))
''' Output:
[144, 72, 48, 24, 36]
[0, 0, 0, 24, 0]
[0, 0, 0, 0, 0]
'''
```

I don’t think the extra overhead is too great, as any zeros are found incidentally with the creation of the product.