54 lines
1.5 KiB
Java
54 lines
1.5 KiB
Java
package DynamicProgramming;
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import java.util.Arrays;
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/**
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* Recursive Solution for 0-1 knapsack with memoization
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*/
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public class KnapsackMemoization {
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private static int[][] t;
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// Returns the maximum value that can
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// be put in a knapsack of capacity W
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public static int knapsack(int[] wt, int[] value, int W, int n) {
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if(t[n][W] != -1) {
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return t[n][W];
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}
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if (n == 0 || W == 0) {
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return 0;
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}
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if (wt[n - 1] <= W) {
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t[n-1][W-wt[n-1]] = knapsack(wt, value, W - wt[n - 1], n - 1);
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// Include item in the bag. In that case add the value of the item and call for the remaining items
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int tmp1 = value[n - 1] + t[n-1][W-wt[n-1]];
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// Don't include the nth item in the bag anl call for remaining item without reducing the weight
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int tmp2 = knapsack(wt, value, W, n - 1);
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t[n-1][W] = tmp2;
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// include the larger one
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int tmp = tmp1 > tmp2 ? tmp1 : tmp2;
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t[n][W] = tmp;
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return tmp;
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// If Weight for the item is more than the desired weight then don't include it
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// Call for rest of the n-1 items
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} else if (wt[n - 1] > W) {
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t[n][W] = knapsack(wt, value, W, n - 1);
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return t[n][W];
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}
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return -1;
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}
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// Driver code
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public static void main(String args[]) {
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int[] wt = {1, 3, 4, 5};
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int[] value = {1, 4, 5, 7};
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int W = 10;
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t = new int[wt.length+1][W+1];
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Arrays.stream(t).forEach(a -> Arrays.fill(a, -1));
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int res = knapsack(wt, value, W, wt.length);
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System.out.println("Maximum knapsack value " + res);
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}
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}
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