Find all differences in an array in O(n)

A first thought is that you aren't using the fact that the array is sorted. Let's assume it's in increasing order (decreasing can be handled analogously).

We can also use the fact that the differences telescope (i>j):

a_i - a_j = (a_i - a_(i-1)) + (a_(i-1) - a_(i-2)) + ... + (a_(j+1) - a_j)

Now build a new sequence, call it s, that has the simple difference, meaning (a_i - a_(i-1)). This takes only one pass (O(n)) to do, and you may as well skip over repeats, meaning skip a_i if a_i = a_(i+1).

All possible differences a_i-a_j with i>j are of the form s_i + s_(i+1) + ... + s_(j+1). So maybe if you count that as having found them, then you did it in O(n) time. To print them, however, may take as many as n(n-1)/2 calls, and that's definitely O(n^2).

For example for an array with the elements {2¹, 2², ..., 2ⁿ} there are n⋅(n-1)/2 possible differences, and no two of them are equal. So there are O(n²) differences.

Since you have to enumerate all of them, you also need at least O(n²) time.