Given weights and values of n items, put these items in a knapsack of capacity W to get the maximum total value in the knapsack. In other words, given two integer arrays val[0..n-1] and wt[0..n-1] which represent values and weights associated with n items respectively. Also given an integer W which represents knapsack capacity, find out the items such that sum of the weights of those items of given subset is smaller than or equal to W. You cannot break an item, either pick the complete item, or don’t pick it (0-1 property).

**Prerequisite :** 0/1 Knapsack

Examples :

Input : val[] = {60, 100, 120}; wt[] = {10, 20, 30}; W = 50; Output : 220 //maximum value that can be obtained 30 20 //weights 20 and 30 are included. Input : val[] = {40, 100, 50, 60}; wt[] = {20, 10, 40, 30}; W = 60; Output : 200 30 20 10

**Approach :**

Let val[] = {1, 4, 5, 7}, wt[] = {1, 3, 4, 5}

W = 7.

The 2d knapsack table will look like :

Start backtracking from K[n][W].Here K[n][W] is 9.

Since, this value comes from the top (shown by grey arrow), the item in this row is not included. Go vertically upward in the table without including this in the knapsack. Now, this value K[n-1][W] which is 9 doesn’t come from the top which means the item in this row is included and go vertically up and then left by the weight of the included item ( shown by black arrow). Continuing this process include weights 3 and 4 with total value 9 in the knapsack.

## C++

// CPP code for Dynamic Programming based // solution for 0-1 Knapsack problem #include <bits/stdc++.h> // A utility function that returns maximum of two integers int max(int a, int b) { return (a > b) ? a : b; } // Prints the items which are put in a knapsack of capacity W void printknapSack(int W, int wt[], int val[], int n) { int i, w; int K[n + 1][W + 1]; // Build table K[][] in bottom up manner for (i = 0; i <= n; i++) { for (w = 0; w <= W; w++) { if (i == 0 || w == 0) K[i][w] = 0; else if (wt[i - 1] <= w) K[i][w] = max(val[i - 1] + K[i - 1][w - wt[i - 1]], K[i - 1][w]); else K[i][w] = K[i - 1][w]; } } // stores the result of Knapsack int res = K[n][W]; printf("%d\n", res); w = W; for (i = n; i > 0 && res > 0; i--) { // either the result comes from the top // (K[i-1][w]) or from (val[i-1] + K[i-1] // [w-wt[i-1]]) as in Knapsack table. If // it comes from the latter one/ it means // the item is included. if (res == K[i - 1][w]) continue; else { // This item is included. printf("%d ", wt[i - 1]); // Since this weight is included its // value is deducted res = res - val[i - 1]; w = w - wt[i - 1]; } } } // Driver code int main() { int val[] = { 60, 100, 120 }; int wt[] = { 10, 20, 30 }; int W = 50; int n = sizeof(val) / sizeof(val[0]); printknapSack(W, wt, val, n); return 0; }

## Java

// Java code for Dynamic Programming based // solution for 0-1 Knapsack problem class GFG { // A utility function that returns // maximum of two integers static int max(int a, int b) { return (a > b) ? a : b; } // Prints the items which are put // in a knapsack of capacity W static void printknapSack(int W, int wt[], int val[], int n) { int i, w; int K[][] = new int[n + 1][W + 1]; // Build table K[][] in bottom up manner for (i = 0; i <= n; i++) { for (w = 0; w <= W; w++) { if (i == 0 || w == 0) K[i][w] = 0; else if (wt[i - 1] <= w) K[i][w] = Math.max(val[i - 1] + K[i - 1][w - wt[i - 1]], K[i - 1][w]); else K[i][w] = K[i - 1][w]; } } // stores the result of Knapsack int res = K[n][W]; System.out.println(res); w = W; for (i = n; i > 0 && res > 0; i--) { // either the result comes from the top // (K[i-1][w]) or from (val[i-1] + K[i-1] // [w-wt[i-1]]) as in Knapsack table. If // it comes from the latter one/ it means // the item is included. if (res == K[i - 1][w]) continue; else { // This item is included. System.out.print(wt[i - 1] + " "); // Since this weight is included its // value is deducted res = res - val[i - 1]; w = w - wt[i - 1]; } } } // Driver code public static void main(String arg[]) { int val[] = { 60, 100, 120 }; int wt[] = { 10, 20, 30 }; int W = 50; int n = val.length; printknapSack(W, wt, val, n); } } // This code is contributed by Anant Agarwal.

## Python3

# Python3 code for Dynamic Programming # based solution for 0-1 Knapsack problem # Prints the items which are put in a # knapsack of capacity W def printknapSack(W, wt, val, n): K = [[0 for w in range(W + 1)] for i in range(n + 1)] # Build table K[][] in bottom # up manner for i in range(n + 1): for w in range(W + 1): if i == 0 or w == 0: K[i][w] = 0 elif wt[i - 1] <= w: K[i][w] = max(val[i - 1] + K[i - 1][w - wt[i - 1]], K[i - 1][w]) else: K[i][w] = K[i - 1][w] # stores the result of Knapsack res = K[n][W] print(res) w = W for i in range(n, 0, -1): if res <= 0: break # either the result comes from the # top (K[i-1][w]) or from (val[i-1] # + K[i-1] [w-wt[i-1]]) as in Knapsack # table. If it comes from the latter # one/ it means the item is included. if res == K[i - 1][w]: continue else: # This item is included. print(wt[i - 1]) # Since this weight is included # its value is deducted res = res - val[i - 1] w = w - wt[i - 1] # Driver code val = [ 60, 100, 120 ] wt = [ 10, 20, 30 ] W = 50 n = len(val) printknapSack(W, wt, val, n) # This code is contributed by Aryan Garg.

**Output:**

220 30 20

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