Question
Given a list of items with values and weights, as well as a max weight, find the maximum value you can generate from items where the sum of the weights is less than the max.
eg.
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items = {(w:1, v:6), (w:2, v:10), (w:3, v:12)} maxWeight = 5 knapsack(items, maxWeight) = 22 |
Once you think that you’ve solved the problem, click below to see the solution.
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Solution
How was that problem? You can check out the solution in the video below.
Here is the source code for the solution shown in the video (Github):
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// Item class public static class Item { int weight; int value; public Item(int weight, int value) { this.weight = weight; this.value = value; } } // Recursively check every combination of items by traversing list of items // and either including or excluding each item public static int naiveKnapsack(Item[] items, int W) { return naiveKnapsack(items, W, 0); } // Overloaded recursive function for naiveKnapsack private static int naiveKnapsack(Item[] items, int W, int i) { // Return when we reach the end of the list if (i == items.length) return 0; // If item is heavier than remaining weight, skip item if (W - items[i].weight < 0) return naiveKnapsack(items, W, i+1); // Try both including and excluding the current item return Math.max(naiveKnapsack(items, W - items[i].weight, i+1) + items[i].value, naiveKnapsack(items, W, i+1)); } // Recursive solution that uses a cache to improve performance public static int topDownKnapsack(Item[] items, int W) { // Map: i -> W -> value // Use a map instead of array because the data could be very sparse Map<Integer, Map<Integer, Integer>> cache = new HashMap<Integer, Map<Integer, Integer>>(); return topDownKnapsack(items, W, 0, cache); } // Overloaded recursive function for topDownKnapsack private static int topDownKnapsack(Item[] items, int W, int i, Map<Integer, Map<Integer, Integer>> cache) { // Return when we reach the end of the list if (i == items.length) return 0; // Check the cache and return value if we get a hit if (!cache.containsKey(i)) cache.put(i, new HashMap<Integer, Integer>()); Integer cached = cache.get(i).get(W); if (cached != null) return cached; // If item is heavier than remaining weight, skip item if (W - items[i].weight < 0) return topDownKnapsack(items, W, i+1, cache); // Try both including and excluding the current item int toReturn = Math.max(topDownKnapsack(items, W - items[i].weight, i+1, cache) + items[i].value, topDownKnapsack(items, W, i+1, cache)); cache.get(i).put(W, toReturn); return toReturn; } // Iterative dynamic programming solution public static int bottomUpKnapsack(Item[] items, int W) { // cache[i][j] = max value for the first i items with a max weight of j int[][] cache = new int[items.length + 1][W + 1]; for (int i = 1; i <= items.length; i++) { for (int j = 0; j <= W; j++) { // If including item[i-1] would exceed max weight j, don't // include the item. Otherwise take the max value of including // or excluding the item if (items[i-1].weight > j) cache[i][j] = cache[i-1][j]; else cache[i][j] = Math.max(cache[i-1][j], cache[i-1][j-items[i-1].weight] + items[i-1].value); } } return cache[items.length][W]; } |
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