I knew this problem but froze for a second trying to remember if there was a DP solution I was supposed to use.
Clarify that the problem allows unlimited transactions with the constraint that you must sell before buying again. Then explain that the optimal strategy is to capture every upward price movement by buying at each local minimum and selling at the next local maximum, which simplifies to summing all positive differences between consecutive days. Finally, present the greedy algorithm with O(n) time and O(1) space, and discuss edge cases.
Pro tip: Emphasize that the greedy approach is not just a heuristic but provably optimal because any transaction spanning multiple days can be decomposed into a series of daily transactions with the same total profit. This demonstrates deep understanding and avoids overcomplicating with dynamic programming.
Confirm that you can buy and sell multiple times, but must sell before buying again, and that you start with no stock. Ask if you can hold at most one share at a time.
Explain that to maximize profit, you should buy at every local minimum and sell at the next local maximum. This is equivalent to capturing all positive day-to-day price increases.
Show that summing all positive differences between consecutive days yields the maximum profit. Provide a simple loop that iterates through the array and adds max(0, prices[i] - prices[i-1]).
State that the algorithm runs in O(n) time and O(1) space, which is optimal since you must examine each price at least once.
Mention edge cases like empty array, single element, or strictly decreasing prices (profit 0). Optionally, discuss how the solution changes if transaction fees or cooldown periods are added.
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