Let's talk about finding the minimum (or maximum only).
If $n$ is the size of the list (not sorted), you will have $n-1$ comparison and $n$ access to the list (if you store the value).
This is due to the fact that numbers are not sorted. So you have to compare all numbers in the given list.
The first with the last, then the minimum of them with the 2nd, the minimum of them etc.
You have to compare the local minimum with all unchecked values otherwise you will miss the results.
Now, you can reduce this number by bounding the list. If you know in advance what is the minimum and/or the maximum (for instance all naturals numbers). You know in advance that if you find a 0, it will be automatically the minimum number in the list. No need to continue checking.
So, in the unsorted list, there is no more efficient way to compare all numbers with the latest minimum found. Or you will make presumption and you will probably get it wrong.
Also, check this good article on finding the minimum and maximum of an array.