Interview Question for Software Engineering at Palantir Technologies:

Dynamic Program. If T time steps are given then can be solved in O(T) time. Create two arrays of length T: Buy[i] = - V[i] + max(Sell[0:i-1]), Sell[i] = V[i] + max(Buy[0:i-1]). i = 1...T-1
Buy[0] = -V[0], Sell[0] = 0
Max operation can be linearized by storing a value.

This can be done even faster and simpler. In the example where you can't short, you should keep track of the minimum value seen so far (working_min). Subtract this from the next element in the list and keep track of the maximum of this difference (working_maxdiff). For each step through the array, check 2 things:
1) if array[i] < working_min, working_min = array[i]
2) if array[i] - working_min > working_maxdiff, then working_maxdiff = array[i] - working_min

If you are allowed to short, all you need is to simultaneously search for the min and max. Subtract them at the end.