# My solution

here’s my solution

``````def same_values(lst1, lst2):
same = []
i = 0

if len(lst1) == len(lst2):
for num in lst1:
if lst1[i] == lst2[i]:
same.append(i)
i = i + 1

return same
``````

Sorry - didn’t realize that’s a faux pas. Just wanted to document one method of completion, since I like looking at how others solve.

I can obscure the solutions.

Yes, that would be good idea. Please do so.

How to obscure solutions?

Use `[details="name"]` to create a show/hide block. Close it with `[/details]`.

Use `[spoiler]` to convert code to a PNG that cannot be copied. Close it with `[/spoiler]`

is_odd
``````def is_odd(x):
return x % 2 and True
``````
```
[details="is_odd"]
[spoiler]
```
def is_odd(x):
return x % 2 and True
```
[/spoiler]
[/details]
```
1 Like

Please critique my solution. After a lot of tinkering, I managed to get it to work in a list comprehension:

``````def same_values(lst1, lst2):
return [lst1.index(x) for x, y in zip(lst1, lst2) if x == y]
``````
``````list.index()
``````

will always give the index of the first instance in the list. It won’t reach any other duplicates.