Sometimes you want to query the small element of a set of elements and use `sorted(elements)[0]` without a second thought. However, this is not optimal, since the entire list must be sorted first `O(n log n)`. It would be better to use the `min(elements)` method provided for this purpose `O(n)`.
Furthermore `sorted(elements)[::-1]` is not very efficient, because it would be better to use `sorted(elements, reverse=True)` to save the slice operation.
**TLDR: using `sorted(elements)[0]` is slow and can be replaced with `min(elements)`.**
I stumbled across these code snippets while playing around with CodeQL (see https://lgtm.com/query/4148064474379348546/).
Pull Request resolved: https://github.com/pytorch/pytorch/pull/86995
Approved by: https://github.com/jansel
