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Remove duplicate KnapsackMemoization (#3769)
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@ -1,51 +1,52 @@
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package com.thealgorithms.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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* This method is basically an extension to the recursive approach so that we
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* can overcome the problem of calculating redundant cases and thus increased
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* complexity. We can solve this problem by simply creating a 2-D array that can
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* store a particular state (n, w) if we get it the first time.
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*/
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public class KnapsackMemoization {
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private static int[][] t;
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int knapSack(int W, int wt[], int val[], int N) {
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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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// Declare the table dynamically
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int dp[][] = new int[N + 1][W + 1];
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// Loop to initially filled the
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// table with -1
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for (int i = 0; i < N + 1; i++) {
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for (int j = 0; j < W + 1; j++) {
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dp[i][j] = -1;
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}
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}
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return knapSackRec(W, wt, val, N, dp);
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}
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// Returns the value of maximum profit using Recursive approach
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int knapSackRec(int W, int wt[],
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int val[], int n,
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int[][] dp) {
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// Base condition
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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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if (dp[n][W] != -1) {
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return dp[n][W];
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}
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if (wt[n - 1] > W) {
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// Store the value of function call stack in table
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dp[n][W] = knapSackRec(W, wt, val, n - 1, dp);
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return dp[n][W];
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} else {
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// Return value of table after storing
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return dp[n][W] = Math.max((val[n - 1] + knapSackRec(W - wt[n - 1], wt, val, n - 1, dp)),
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knapSackRec(W, wt, val, n - 1, dp));
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}
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}
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}
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@ -1,68 +0,0 @@
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package com.thealgorithms.dynamicprogramming;
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// Here is the top-down approach of
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// dynamic programming
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public class MemoizationTechniqueKnapsack {
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// A utility function that returns
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// maximum of two integers
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static int max(int a, int b) {
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return (a > b) ? a : b;
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}
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// Returns the value of maximum profit
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static int knapSackRec(int W, int wt[], int val[], int n, int[][] dp) {
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// Base condition
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if (n == 0 || W == 0) {
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return 0;
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}
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if (dp[n][W] != -1) {
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return dp[n][W];
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}
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if (
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wt[n - 1] > W
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) { // stack in table before return // Store the value of function call
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return dp[n][W] = knapSackRec(W, wt, val, n - 1, dp);
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} else { // Return value of table after storing
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return (
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dp[n][W] =
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max(
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(
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val[n - 1] +
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knapSackRec(W - wt[n - 1], wt, val, n - 1, dp)
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),
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knapSackRec(W, wt, val, n - 1, dp)
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)
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);
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}
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}
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static int knapSack(int W, int wt[], int val[], int N) {
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// Declare the table dynamically
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int dp[][] = new int[N + 1][W + 1];
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// Loop to initially filled the
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// table with -1
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for (int i = 0; i < N + 1; i++) {
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for (int j = 0; j < W + 1; j++) {
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dp[i][j] = -1;
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}
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}
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return knapSackRec(W, wt, val, N, dp);
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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 val[] = { 60, 100, 120 };
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int wt[] = { 10, 20, 30 };
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int W = 50;
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int N = val.length;
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System.out.println(knapSack(W, wt, val, N));
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}
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}
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