algorithm_fridays_week04.py

from functools import reduce

def get_products(nums):
  products = []

  if nums == None or len(nums) == 0 or len(nums) == 1:
    return products

  for i, el in enumerate(nums):
    nums_to_multiply = nums[:i] + nums[i+1:]
    product = reduce(lambda x, y: x * y, nums_to_multiply)
    products.append(product)
  
  return products

if __name__ == '__main__':
  nums = [4, 5, 10, 2]
  products = get_products(nums)
  print(products)

Thanks a lot for your feedback @meekg33k.

Yeah, it's equivalent to a nested loop, because reduce takes about O(n) since it iterates through all the elements, even though it's n-1 elements in this case, that's still ~n. And then I'm also iterating through each element as well. So time complexity is O(n^2), because for each element I go through in my iteration, reduce goes through all other elements.

I initially thought I could do it without having to create a new array, but I would be modifying the existing one, making it impossible to get the subsequent products as I proceed. So I had to create a new array to hold each product to avoid mutation of the original. This implies double the memory space, so not very memory efficient if we consider that nums is a very large array. I'm already excited to see your incoming solution and how you handle this.

Also, I store the other elements to be multiplied in a temporary array as well during my iteration (I could have passed directly to reduce, but it wouldn't be easily readable), but each iteration does away the space every time, so that was acceptable to me.

Thanks again, I really appreciate.