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| 1 | +package dynamicprogramming; |
| 2 | +/** |
| 3 | + * Implementation of the 0/1 Knapsack Problem using Dynamic Programming. |
| 4 | + * |
| 5 | + * Given weights and values of n items, put these items in a knapsack of capacity W |
| 6 | + * to get the maximum total value in the knapsack. |
| 7 | + */ |
| 8 | +public class Knapsack { |
| 9 | + |
| 10 | + /** |
| 11 | + * Returns the maximum value that can be put in a knapsack of capacity W. |
| 12 | + * |
| 13 | + * @param W capacity of the knapsack |
| 14 | + * @param wt array of weights of the items |
| 15 | + * @param val array of values of the items |
| 16 | + * @param n number of items |
| 17 | + * @return maximum value achievable with given capacity |
| 18 | + */ |
| 19 | + public static int knapSack(int W, int[] wt, int[] val, int n) { |
| 20 | + int[][] dp = new int[n + 1][W + 1]; |
| 21 | + |
| 22 | + // Build table dp[][] in bottom-up manner |
| 23 | + for (int i = 0; i <= n; i++) { |
| 24 | + for (int w = 0; w <= W; w++) { |
| 25 | + if (i == 0 || w == 0) { |
| 26 | + dp[i][w] = 0; // Base case: no items or zero capacity |
| 27 | + } else if (wt[i - 1] <= w) { |
| 28 | + // Include the item or exclude it, choose max value |
| 29 | + dp[i][w] = Math.max(val[i - 1] + dp[i - 1][w - wt[i - 1]], |
| 30 | + dp[i - 1][w]); |
| 31 | + } else { |
| 32 | + // Item can't be included because it weighs more than capacity |
| 33 | + dp[i][w] = dp[i - 1][w]; |
| 34 | + } |
| 35 | + } |
| 36 | + } |
| 37 | + |
| 38 | + return dp[n][W]; |
| 39 | + } |
| 40 | + |
| 41 | + // Example usage and simple test case |
| 42 | + public static void main(String[] args) { |
| 43 | + int[] values = { 60, 100, 120 }; |
| 44 | + int[] weights = { 10, 20, 30 }; |
| 45 | + int capacity = 50; |
| 46 | + int n = values.length; |
| 47 | + |
| 48 | + int maxValue = knapSack(capacity, weights, values, n); |
| 49 | + System.out.println("Maximum value in Knapsack = " + maxValue); |
| 50 | + // Expected output: 220 |
| 51 | + } |
| 52 | +} |
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