#toyprogrammingchallenge

Okay, here is another freebie :) I will put in a real one before the end of the weekend.

Basically, you are given a list from which you need to create a new list where each element is the product of all the "OTHER" elements.

I found a few interesting corner cases.

I challenge you to give it a try!

Read the challenge here:

https://git.qoto.org/Absinthe/productnot/blob/master/README.md

My attempt is checked into that repo as well.

@Absinthe Here is my #toyprogrammingchallenge no-division solution:

```
#!/usr/bin/env python
from collections import deque
from functools import reduce
from random import randint

def get_product_and_rotate(d):
"""Gets product of all but the first items in the queue"""
prod = reduce(lambda a, b: a * b, list(d)[1:])
d.rotate(-1) # Rotates left so that the next
return prod

if __name__ == "__main__":
n_items = randint(2, 10)
input_list = [randint(0, 9) for _ in range(n_items)]
print(input_list)
product_queue = deque(input_list)
print([get_product_and_rotate(product_queue) for _ in input_list])
```

Uses `reduce` for the multiplication, and uses a `deque` (double-ended queue) to rotate the list. I've read that using a deque is actually more efficient than just taking slices out of the list.

@zingbretsen What about this... as a one liner:

[reduce(lambda a, b: a *b, [ y for x, y in enumerate(list) if x != idx]) for idx, _ in enumerate(list)]

@Absinthe looks right to me

@zingbretsen This works...

print("One Liner no division")
print([reduce(lambda a, b: a *b,
[ y for x, y in enumerate(random_list) if x != idx])
for idx, _ in enumerate(random_list)])

It's a one liner, except I am sticking to 80 columns so I split it a few times.. still calling it a one liner :)

@Absinthe looks good.

I'll try to do something with the dynamic programming approach tomorrow or Monday.

For something like multiplication, I don't think it will make much of a difference in the execution time, but for a more expensive operation it would