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Commit 2bbe12b

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Add divide and conquer example: best time to buy and sell stocks (trekhleb#612)
* Add divide and conquer example: best time to buy and sell stocks. * Add divide and conquer example: best time to buy and sell stocks. * Add dynamic programming version.
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‎README.md

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@@ -129,9 +129,9 @@ a set of rules that precisely define a sequence of operations.
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* `B` [Depth-First Search](src/algorithms/graph/depth-first-search) (DFS)
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* `B` [Breadth-First Search](src/algorithms/graph/breadth-first-search) (BFS)
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* `B` [Kruskal’s Algorithm](src/algorithms/graph/kruskal) - finding Minimum Spanning Tree (MST) for weighted undirected graph
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* `A` [Dijkstra Algorithm](src/algorithms/graph/dijkstra) - finding shortest paths to all graph vertices from single vertex
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* `A` [Bellman-Ford Algorithm](src/algorithms/graph/bellman-ford) - finding shortest paths to all graph vertices from single vertex
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* `A` [Floyd-Warshall Algorithm](src/algorithms/graph/floyd-warshall) - find shortest paths between all pairs of vertices
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* `A` [Dijkstra Algorithm](src/algorithms/graph/dijkstra) - finding the shortest paths to all graph vertices from single vertex
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* `A` [Bellman-Ford Algorithm](src/algorithms/graph/bellman-ford) - finding the shortest paths to all graph vertices from single vertex
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* `A` [Floyd-Warshall Algorithm](src/algorithms/graph/floyd-warshall) - find the shortest paths between all pairs of vertices
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* `A` [Detect Cycle](src/algorithms/graph/detect-cycle) - for both directed and undirected graphs (DFS and Disjoint Set based versions)
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* `A` [Prim’s Algorithm](src/algorithms/graph/prim) - finding Minimum Spanning Tree (MST) for weighted undirected graph
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* `A` [Topological Sorting](src/algorithms/graph/topological-sorting) - DFS method
@@ -157,6 +157,7 @@ a set of rules that precisely define a sequence of operations.
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* `B` [Unique Paths](src/algorithms/uncategorized/unique-paths) - backtracking, dynamic programming and Pascal's Triangle based examples
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* `B` [Rain Terraces](src/algorithms/uncategorized/rain-terraces) - trapping rain water problem (dynamic programming and brute force versions)
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* `B` [Recursive Staircase](src/algorithms/uncategorized/recursive-staircase) - count the number of ways to reach to the top (4 solutions)
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* `B` [Best Time To Buy Sell Stocks](src/algorithms/uncategorized/best-time-to-buy-sell-stocks) - divide and conquer and one-pass examples
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* `A` [N-Queens Problem](src/algorithms/uncategorized/n-queens)
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* `A` [Knight's Tour](src/algorithms/uncategorized/knight-tour)
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* **Greedy** - choose the best option at the current time, without any consideration for the future
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* `B` [Jump Game](src/algorithms/uncategorized/jump-game)
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* `A` [Unbound Knapsack Problem](src/algorithms/sets/knapsack-problem)
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* `A` [Dijkstra Algorithm](src/algorithms/graph/dijkstra) - finding shortest path to all graph vertices
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* `A` [Dijkstra Algorithm](src/algorithms/graph/dijkstra) - finding the shortest path to all graph vertices
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* `A` [Prim’s Algorithm](src/algorithms/graph/prim) - finding Minimum Spanning Tree (MST) for weighted undirected graph
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* `A` [Kruskal’s Algorithm](src/algorithms/graph/kruskal) - finding Minimum Spanning Tree (MST) for weighted undirected graph
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* **Divide and Conquer** - divide the problem into smaller parts and then solve those parts
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* `B` [Matrices](src/algorithms/math/matrix) - generating and traversing the matrices of different shapes
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* `B` [Jump Game](src/algorithms/uncategorized/jump-game)
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* `B` [Fast Powering](src/algorithms/math/fast-powering)
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* `B` [Best Time To Buy Sell Stocks](src/algorithms/uncategorized/best-time-to-buy-sell-stocks) - divide and conquer and one-pass examples
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* `A` [Permutations](src/algorithms/sets/permutations) (with and without repetitions)
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* `A` [Combinations](src/algorithms/sets/combinations) (with and without repetitions)
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* **Dynamic Programming** - build up a solution using previously found sub-solutions
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* `A` [0/1 Knapsack Problem](src/algorithms/sets/knapsack-problem)
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* `A` [Integer Partition](src/algorithms/math/integer-partition)
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* `A` [Maximum Subarray](src/algorithms/sets/maximum-subarray)
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* `A` [Bellman-Ford Algorithm](src/algorithms/graph/bellman-ford) - finding shortest path to all graph vertices
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* `A` [Floyd-Warshall Algorithm](src/algorithms/graph/floyd-warshall) - find shortest paths between all pairs of vertices
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* `A` [Bellman-Ford Algorithm](src/algorithms/graph/bellman-ford) - finding the shortest path to all graph vertices
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* `A` [Floyd-Warshall Algorithm](src/algorithms/graph/floyd-warshall) - find the shortest paths between all pairs of vertices
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* `A` [Regular Expression Matching](src/algorithms/string/regular-expression-matching)
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* **Backtracking** - similarly to brute force, try to generate all possible solutions, but each time you generate next solution you test
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if it satisfies all conditions, and only then continue generating subsequent solutions. Otherwise, backtrack, and go on a
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# Best Time to Buy and Sell Stock
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## Task Description
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Say you have an array prices for which the `i`-th element is the price of a given stock on day `i`.
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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).
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> Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
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**Example #1**
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```
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Input: [7, 1, 5, 3, 6, 4]
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Output: 7
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```
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_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`.
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**Example #2**
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```
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Input: [1, 2, 3, 4, 5]
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Output: 4
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```
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_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.
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**Example #3**
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```
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Input: [7, 6, 4, 3, 1]
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Output: 0
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```
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_Explanation:_ In this case, no transaction is done, i.e. max `profit = 0`.
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## Possible Solutions
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### Divide and conquer approach `O(2^n)`
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We may try **all** combinations of buying and selling and find out the most profitable one by applying _divide and conquer approach_.
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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:
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1. _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])`.
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2. _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])`.
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The overall profit would be equal to → `overalProfit = Max(keepProfit, buySellProfit)`.
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As you can see the `profit([6, 4, 3, 1])` task is being solved in the same recursive manner.
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> See the full code example in [dqBestTimeToBuySellStocks.js](dqBestTimeToBuySellStocks.js)
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#### Time Complexity
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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)`.
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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`!
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#### Additional Space Complexity
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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)`
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## Peak Valley Approach `O(n)`
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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
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![Peak Valley Approach](https://leetcode.com/media/original_images/122_maxprofit_1.PNG)
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_Image source: [LeetCode](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/solution/)_
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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).
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> See the full code example in [peakvalleyBestTimeToBuySellStocks.js](peakvalleyBestTimeToBuySellStocks.js)
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#### Time Complexity
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Since the algorithm requires only one pass through the prices array, the time complexity would equal `O(n)`.
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#### Additional Space Complexity
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Except of the prices array itself the algorithm consumes the constant amount of memory. Thus, additional space complexity is `O(1)`.
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## Accumulator Approach `O(n)`
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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]`:
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![Simple One Pass](https://leetcode.com/media/original_images/122_maxprofit_2.PNG)
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_Image source: [LeetCode](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/solution/)_
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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,
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> See the full code example in [accumulatorBestTimeToBuySellStocks.js](accumulatorBestTimeToBuySellStocks.js)
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#### Time Complexity
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Since the algorithm requires only one pass through the prices array, the time complexity would equal `O(n)`.
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#### Additional Space Complexity
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Except of the prices array itself the algorithm consumes the constant amount of memory. Thus, additional space complexity is `O(1)`.
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## References
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- [Best Time to Buy and Sell Stock on LeetCode](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/)
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import accumulatorBestTimeToBuySellStocks from '../accumulatorBestTimeToBuySellStocks';
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describe('accumulatorBestTimeToBuySellStocks', () => {
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it('should find the best time to buy and sell stocks', () => {
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let visit;
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expect(accumulatorBestTimeToBuySellStocks([1, 5])).toEqual(4);
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visit = jest.fn();
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expect(accumulatorBestTimeToBuySellStocks([1], visit)).toEqual(0);
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expect(visit).toHaveBeenCalledTimes(1);
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visit = jest.fn();
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expect(accumulatorBestTimeToBuySellStocks([1, 5], visit)).toEqual(4);
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expect(visit).toHaveBeenCalledTimes(2);
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visit = jest.fn();
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expect(accumulatorBestTimeToBuySellStocks([5, 1], visit)).toEqual(0);
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expect(visit).toHaveBeenCalledTimes(2);
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visit = jest.fn();
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expect(accumulatorBestTimeToBuySellStocks([1, 5, 10], visit)).toEqual(9);
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expect(visit).toHaveBeenCalledTimes(3);
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visit = jest.fn();
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expect(accumulatorBestTimeToBuySellStocks([10, 1, 5, 20, 15, 21], visit)).toEqual(25);
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expect(visit).toHaveBeenCalledTimes(6);
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visit = jest.fn();
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expect(accumulatorBestTimeToBuySellStocks([7, 1, 5, 3, 6, 4], visit)).toEqual(7);
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expect(visit).toHaveBeenCalledTimes(6);
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visit = jest.fn();
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expect(accumulatorBestTimeToBuySellStocks([1, 2, 3, 4, 5], visit)).toEqual(4);
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expect(visit).toHaveBeenCalledTimes(5);
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visit = jest.fn();
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expect(accumulatorBestTimeToBuySellStocks([7, 6, 4, 3, 1], visit)).toEqual(0);
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expect(visit).toHaveBeenCalledTimes(5);
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visit = jest.fn();
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expect(accumulatorBestTimeToBuySellStocks(
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[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20],
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visit,
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)).toEqual(19);
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expect(visit).toHaveBeenCalledTimes(20);
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});
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});
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import dpBestTimeToBuySellStocks from '../dpBestTimeToBuySellStocks';
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describe('dpBestTimeToBuySellStocks', () => {
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it('should find the best time to buy and sell stocks', () => {
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let visit;
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expect(dpBestTimeToBuySellStocks([1, 5])).toEqual(4);
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visit = jest.fn();
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expect(dpBestTimeToBuySellStocks([1], visit)).toEqual(0);
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expect(visit).toHaveBeenCalledTimes(1);
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visit = jest.fn();
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expect(dpBestTimeToBuySellStocks([1, 5], visit)).toEqual(4);
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expect(visit).toHaveBeenCalledTimes(2);
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visit = jest.fn();
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expect(dpBestTimeToBuySellStocks([5, 1], visit)).toEqual(0);
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expect(visit).toHaveBeenCalledTimes(2);
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visit = jest.fn();
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expect(dpBestTimeToBuySellStocks([1, 5, 10], visit)).toEqual(9);
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expect(visit).toHaveBeenCalledTimes(3);
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visit = jest.fn();
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expect(dpBestTimeToBuySellStocks([10, 1, 5, 20, 15, 21], visit)).toEqual(25);
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expect(visit).toHaveBeenCalledTimes(6);
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visit = jest.fn();
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expect(dpBestTimeToBuySellStocks([7, 1, 5, 3, 6, 4], visit)).toEqual(7);
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expect(visit).toHaveBeenCalledTimes(6);
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visit = jest.fn();
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expect(dpBestTimeToBuySellStocks([1, 2, 3, 4, 5], visit)).toEqual(4);
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expect(visit).toHaveBeenCalledTimes(5);
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visit = jest.fn();
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expect(dpBestTimeToBuySellStocks([7, 6, 4, 3, 1], visit)).toEqual(0);
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expect(visit).toHaveBeenCalledTimes(5);
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visit = jest.fn();
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expect(dpBestTimeToBuySellStocks(
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[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20],
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visit,
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)).toEqual(19);
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expect(visit).toHaveBeenCalledTimes(20);
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});
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});
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import dqBestTimeToBuySellStocks from '../dqBestTimeToBuySellStocks';
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describe('dqBestTimeToBuySellStocks', () => {
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it('should find the best time to buy and sell stocks', () => {
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let visit;
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expect(dqBestTimeToBuySellStocks([1, 5])).toEqual(4);
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visit = jest.fn();
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expect(dqBestTimeToBuySellStocks([1], visit)).toEqual(0);
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expect(visit).toHaveBeenCalledTimes(3);
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visit = jest.fn();
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expect(dqBestTimeToBuySellStocks([1, 5], visit)).toEqual(4);
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expect(visit).toHaveBeenCalledTimes(7);
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visit = jest.fn();
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expect(dqBestTimeToBuySellStocks([5, 1], visit)).toEqual(0);
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expect(visit).toHaveBeenCalledTimes(7);
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visit = jest.fn();
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expect(dqBestTimeToBuySellStocks([1, 5, 10], visit)).toEqual(9);
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expect(visit).toHaveBeenCalledTimes(15);
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visit = jest.fn();
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expect(dqBestTimeToBuySellStocks([10, 1, 5, 20, 15, 21], visit)).toEqual(25);
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expect(visit).toHaveBeenCalledTimes(127);
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visit = jest.fn();
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expect(dqBestTimeToBuySellStocks([7, 1, 5, 3, 6, 4], visit)).toEqual(7);
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expect(visit).toHaveBeenCalledTimes(127);
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visit = jest.fn();
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expect(dqBestTimeToBuySellStocks([1, 2, 3, 4, 5], visit)).toEqual(4);
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expect(visit).toHaveBeenCalledTimes(63);
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visit = jest.fn();
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expect(dqBestTimeToBuySellStocks([7, 6, 4, 3, 1], visit)).toEqual(0);
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expect(visit).toHaveBeenCalledTimes(63);
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visit = jest.fn();
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expect(dqBestTimeToBuySellStocks(
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[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20],
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visit,
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)).toEqual(19);
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expect(visit).toHaveBeenCalledTimes(2097151);
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});
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});
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import peakvalleyBestTimeToBuySellStocks from '../peakvalleyBestTimeToBuySellStocks';
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describe('peakvalleyBestTimeToBuySellStocks', () => {
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it('should find the best time to buy and sell stocks', () => {
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let visit;
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expect(peakvalleyBestTimeToBuySellStocks([1, 5])).toEqual(4);
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visit = jest.fn();
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expect(peakvalleyBestTimeToBuySellStocks([1], visit)).toEqual(0);
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expect(visit).toHaveBeenCalledTimes(1);
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visit = jest.fn();
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expect(peakvalleyBestTimeToBuySellStocks([1, 5], visit)).toEqual(4);
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expect(visit).toHaveBeenCalledTimes(2);
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visit = jest.fn();
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expect(peakvalleyBestTimeToBuySellStocks([5, 1], visit)).toEqual(0);
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expect(visit).toHaveBeenCalledTimes(2);
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visit = jest.fn();
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expect(peakvalleyBestTimeToBuySellStocks([1, 5, 10], visit)).toEqual(9);
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expect(visit).toHaveBeenCalledTimes(3);
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visit = jest.fn();
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expect(peakvalleyBestTimeToBuySellStocks([10, 1, 5, 20, 15, 21], visit)).toEqual(25);
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expect(visit).toHaveBeenCalledTimes(6);
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visit = jest.fn();
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expect(peakvalleyBestTimeToBuySellStocks([7, 1, 5, 3, 6, 4], visit)).toEqual(7);
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expect(visit).toHaveBeenCalledTimes(6);
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visit = jest.fn();
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expect(peakvalleyBestTimeToBuySellStocks([1, 2, 3, 4, 5], visit)).toEqual(4);
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expect(visit).toHaveBeenCalledTimes(5);
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visit = jest.fn();
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expect(peakvalleyBestTimeToBuySellStocks([7, 6, 4, 3, 1], visit)).toEqual(0);
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expect(visit).toHaveBeenCalledTimes(5);
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visit = jest.fn();
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expect(peakvalleyBestTimeToBuySellStocks(
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[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20],
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visit,
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)).toEqual(19);
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expect(visit).toHaveBeenCalledTimes(20);
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});
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});
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/**
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* Finds the maximum profit from selling and buying the stocks.
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* ACCUMULATOR APPROACH.
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*
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* @param {number[]} prices - Array of stock prices, i.e. [7, 6, 4, 3, 1]
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* @param {function(): void} visit - Visiting callback to calculate the number of iterations.
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* @return {number} - The maximum profit
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*/
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const accumulatorBestTimeToBuySellStocks = (prices, visit = () => {}) => {
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visit();
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let profit = 0;
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for (let day = 1; day < prices.length; day += 1) {
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visit();
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// Add the increase of the price from yesterday till today (if there was any) to the profit.
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profit += Math.max(prices[day] - prices[day - 1], 0);
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}
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return profit;
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};
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export default accumulatorBestTimeToBuySellStocks;
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/**
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* Finds the maximum profit from selling and buying the stocks.
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* DYNAMIC PROGRAMMING APPROACH.
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*
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* @param {number[]} prices - Array of stock prices, i.e. [7, 6, 4, 3, 1]
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* @param {function(): void} visit - Visiting callback to calculate the number of iterations.
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* @return {number} - The maximum profit
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*/
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const dpBestTimeToBuySellStocks = (prices, visit = () => {}) => {
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visit();
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let lastBuy = -prices[0];
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let lastSold = 0;
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for (let day = 1; day < prices.length; day += 1) {
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visit();
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const curBuy = Math.max(lastBuy, lastSold - prices[day]);
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const curSold = Math.max(lastSold, lastBuy + prices[day]);
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lastBuy = curBuy;
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lastSold = curSold;
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}
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return lastSold;
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};
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export default dpBestTimeToBuySellStocks;

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