Say you have an array prices for which the i-th element is the price of a given stock on day i.
Find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).
Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
Example #1
Input: [7, 1, 5, 3, 6, 4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4. Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
Example #2
Input: [1, 2, 3, 4, 5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4. Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.
Example #3
Input: [7, 6, 4, 3, 1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
We may try all combinations of buying and selling and find out the most profitable one by applying divide and conquer approach.
Let's say we have an array of prices [7, 6, 4, 3, 1] and we're on the 1st day of trading (at the very beginning of the array). At this point we may say that the overall maximum profit would be the maximum of two following values:
- Option 1: Keep the money → profit would equal to the profit from buying/selling the rest of the stocks →
keepProfit = profit([6, 4, 3, 1]). - Option 2: Buy/sell at current price → profit in this case would equal to the profit from buying/selling the rest of the stocks plus (or minus, depending on whether we're selling or buying) the current stock price →
buySellProfit = -7 + profit([6, 4, 3, 1]).
The overall profit would be equal to → overalProfit = Max(keepProfit, buySellProfit).
As you can see the profit([6, 4, 3, 1]) task is being solved in the same recursive manner.
See the full code example in dqBestTimeToBuySellStocks.js
As you may see, every recursive call will produce 2 more recursive branches. The depth of the recursion will be n (size of prices array) and thus, the time complexity will equal to O(2^n).
As you may see, this is very inefficient. For example for just 20 prices the number of recursive calls will be somewhere close to 2M!
If we avoid cloning the prices array between recursive function calls and will use the array pointer then additional space complexity will be proportional to the depth of the recursion: O(n)
If we plot the prices array (i.e. [7, 1, 5, 3, 6, 4]) we may notice that the points of interest are the consecutive valleys and peaks
Image source: LeetCode
So, if we will track the growing price and will sell the stocks immediately before the price goes down we'll get the maximum profit (remember, we bought the stock in the valley at its low price).
See the full code example in peakvalleyBestTimeToBuySellStocks.js
Since the algorithm requires only one pass through the prices array, the time complexity would equal O(n).
Except of the prices array itself the algorithm consumes the constant amount of memory. Thus, additional space complexity is O(1).
There is even simpler approach exists. Let's say we have the prices array which looks like this [1, 7, 2, 3, 6, 7, 6, 7]:
Image source: LeetCode
You may notice, that we don't even need to keep tracking of a constantly growing price. Instead, we may simply add the price difference for all growing segments of the chart which eventually sums up to the highest possible profit,
See the full code example in accumulatorBestTimeToBuySellStocks.js
Since the algorithm requires only one pass through the prices array, the time complexity would equal O(n).
Except of the prices array itself the algorithm consumes the constant amount of memory. Thus, additional space complexity is O(1).