Last Updated : 30 May, 2024
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Prerequisites: Introduction to Knapsack Problem, its Types and How to solve them
Given N items where each item has some weight and profit associated with it and also given a bag with capacity W, [i.e., the bag can hold at most W weight in it]. The task is to put the items into the bag such that the sum of profits associated with them is the maximum possible.
Note: The constraint here is we can either put an item completely into the bag or cannot put it at all [It is not possible to put a part of an item into the bag].
Examples:
Input: N = 3, W = 4, profit[] = {1, 2, 3}, weight[] = {4, 5, 1}
Output: 3
Explanation: There are two items which have weight less than or equal to 4. If we select the item with weight 4, the possible profit is 1. And if we select the item with weight 1, the possible profit is 3. So the maximum possible profit is 3. Note that we cannot put both the items with weight 4 and 1 together as the capacity of the bag is 4.Input: N = 3, W = 3, profit[] = {1, 2, 3}, weight[] = {4, 5, 6}
Output: 0
Recursion Approach for 0/1 Knapsack Problem:
To solve the problem follow the below idea:
A simple solution is to consider all subsets of items and calculate the total weight and profit of all subsets. Consider the only subsets whose total weight is smaller than W. From all such subsets, pick the subset with maximum profit.
Optimal Substructure: To consider all subsets of items, there can be two cases for every item.
- Case 1: The item is included in the optimal subset.
- Case 2: The item is not included in the optimal set.
Follow the below steps to solve the problem:
The maximum value obtained from ‘N’ items is the max of the following two values.
- Case 1 (include the Nth item): Value of the Nth item plus maximum value obtained by remaining N-1 items and remaining weight i.e. (W-weight of the Nth item).
- Case 2 (exclude the Nth item): Maximum value obtained by N-1 items and W weight.
- If the weight of the ‘Nth‘ item is greater than ‘W’, then the Nth item cannot be included and Case 2 is the only possibility.
Below is the implementation of the above approach:
/* A Naive recursive implementation of 0-1 Knapsack problem */#include <bits/stdc++.h>using namespace std;// A utility function that returns// maximum of two integersint max(int a, int b) { return (a > b) ? a : b; }// Returns the maximum value that// can be put in a knapsack of capacity Wint knapSack(int W, int wt[], int val[], int n){ // Base Case if (n == 0 || W == 0) return 0; // If weight of the nth item is more // than Knapsack capacity W, then // this item cannot be included // in the optimal solution if (wt[n - 1] > W) return knapSack(W, wt, val, n - 1); // Return the maximum of two cases: // (1) nth item included // (2) not included else return max( val[n - 1] + knapSack(W - wt[n - 1], wt, val, n - 1), knapSack(W, wt, val, n - 1));}// Driver codeint main(){ int profit[] = { 60, 100, 120 }; int weight[] = { 10, 20, 30 }; int W = 50; int n = sizeof(profit) / sizeof(profit[0]); cout << knapSack(W, weight, profit, n); return 0;}// This code is contributed by rathbhupendra
/* A Naive recursive implementationof 0-1 Knapsack problem */#include <stdio.h>// A utility function that returns// maximum of two integersint max(int a, int b) { return (a > b) ? a : b; }// Returns the maximum value that can be// put in a knapsack of capacity Wint knapSack(int W, int wt[], int val[], int n){ // Base Case if (n == 0 || W == 0) return 0; // If weight of the nth item is more than // Knapsack capacity W, then this item cannot // be included in the optimal solution if (wt[n - 1] > W) return knapSack(W, wt, val, n - 1); // Return the maximum of two cases: // (1) nth item included // (2) not included else return max( val[n - 1] + knapSack(W - wt[n - 1], wt, val, n - 1), knapSack(W, wt, val, n - 1));}// Driver codeint main(){ int profit[] = { 60, 100, 120 }; int weight[] = { 10, 20, 30 }; int W = 50; int n = sizeof(profit) / sizeof(profit[0]); printf("%d", knapSack(W, weight, profit, n)); return 0;}
/* A Naive recursive implementationof 0-1 Knapsack problem */class Knapsack { // A utility function that returns // maximum of two integers static int max(int a, int b) { return (a > b) ? a : b; } // Returns the maximum value that // can be put in a knapsack of // capacity W static int knapSack(int W, int wt[], int val[], int n) { // Base Case if (n == 0 || W == 0) return 0; // If weight of the nth item is // more than Knapsack capacity W, // then this item cannot be included // in the optimal solution if (wt[n - 1] > W) return knapSack(W, wt, val, n - 1); // Return the maximum of two cases: // (1) nth item included // (2) not included else return max(val[n - 1] + knapSack(W - wt[n - 1], wt, val, n - 1), knapSack(W, wt, val, n - 1)); } // Driver code public static void main(String args[]) { int profit[] = new int[] { 60, 100, 120 }; int weight[] = new int[] { 10, 20, 30 }; int W = 50; int n = profit.length; System.out.println(knapSack(W, weight, profit, n)); }}/*This code is contributed by Rajat Mishra */
# A naive recursive implementation# of 0-1 Knapsack Problem# Returns the maximum value that# can be put in a knapsack of# capacity Wdef knapSack(W, wt, val, n): # Base Case if n == 0 or W == 0: return 0 # If weight of the nth item is # more than Knapsack of capacity W, # then this item cannot be included # in the optimal solution if (wt[n-1] > W): return knapSack(W, wt, val, n-1) # return the maximum of two cases: # (1) nth item included # (2) not included else: return max( val[n-1] + knapSack( W-wt[n-1], wt, val, n-1), knapSack(W, wt, val, n-1))# end of function knapSack# Driver Codeif __name__ == '__main__': profit = [60, 100, 120] weight = [10, 20, 30] W = 50 n = len(profit) print knapSack(W, weight, profit, n)# This code is contributed by Nikhil Kumar Singh
/* A Naive recursive implementation of0-1 Knapsack problem */using System;class GFG { // A utility function that returns // maximum of two integers static int max(int a, int b) { return (a > b) ? a : b; } // Returns the maximum value that can // be put in a knapsack of capacity W static int knapSack(int W, int[] wt, int[] val, int n) { // Base Case if (n == 0 || W == 0) return 0; // If weight of the nth item is // more than Knapsack capacity W, // then this item cannot be // included in the optimal solution if (wt[n - 1] > W) return knapSack(W, wt, val, n - 1); // Return the maximum of two cases: // (1) nth item included // (2) not included else return max(val[n - 1] + knapSack(W - wt[n - 1], wt, val, n - 1), knapSack(W, wt, val, n - 1)); } // Driver code public static void Main() { int[] profit = new int[] { 60, 100, 120 }; int[] weight = new int[] { 10, 20, 30 }; int W = 50; int n = profit.Length; Console.WriteLine(knapSack(W, weight, profit, n)); }}// This code is contributed by Sam007
/* A Naive recursive implementation of 0-1 Knapsack problem */ // A utility function that returns // maximum of two integers function max(a, b) { return (a > b) ? a : b; } // Returns the maximum value that can // be put in a knapsack of capacity W function knapSack(W, wt, val, n) { // Base Case if (n == 0 || W == 0) return 0; // If weight of the nth item is // more than Knapsack capacity W, // then this item cannot be // included in the optimal solution if (wt[n - 1] > W) return knapSack(W, wt, val, n - 1); // Return the maximum of two cases: // (1) nth item included // (2) not included else return max(val[n - 1] + knapSack(W - wt[n - 1], wt, val, n - 1), knapSack(W, wt, val, n - 1)); } let profit = [ 60, 100, 120 ]; let weight = [ 10, 20, 30 ]; let W = 50; let n = profit.length; console.log(knapSack(W, weight, profit, n));
<?php// A Naive recursive implementation// of 0-1 Knapsack problem // Returns the maximum value that// can be put in a knapsack of // capacity Wfunction knapSack($W, $wt, $val, $n){ // Base Case if ($n == 0 || $W == 0) return 0; // If weight of the nth item is // more than Knapsack capacity // W, then this item cannot be // included in the optimal solution if ($wt[$n - 1] > $W) return knapSack($W, $wt, $val, $n - 1); // Return the maximum of two cases: // (1) nth item included // (2) not included else return max($val[$n - 1] + knapSack($W - $wt[$n - 1], $wt, $val, $n - 1), knapSack($W, $wt, $val, $n-1));} // Driver Code $profit = array(60, 100, 120); $weight = array(10, 20, 30); $W = 50; $n = count($profit); echo knapSack($W, $weight, $profit, $n);// This code is contributed by Sam007?>
Output
220
Time Complexity: O(2N)
Auxiliary Space: O(N), Stack space required for recursion
Dynamic Programming Approach for 0/1 Knapsack Problem
Memoization Approach for 0/1 Knapsack Problem:
Note: It should be noted that the above function using recursion computes the same subproblems again and again. See the following recursion tree, K(1, 1) is being evaluated twice.
In the following recursion tree, K() refers to knapSack(). The two parameters indicated in the following recursion tree are n and W.
The recursion tree is for following sample inputs.
weight[] = {1, 1, 1}, W = 2, profit[] = {10, 20, 30}K(3, 2)
/ \
/ \
K(2, 2) K(2, 1)
/ \ / \
/ \ / \
K(1, 2) K(1, 1) K(1, 1) K(1, 0)
/ \ / \ / \
/ \ / \ / \
K(0, 2) K(0, 1) K(0, 1) K(0, 0) K(0, 1) K(0, 0)Recursion tree for Knapsack capacity 2 units and 3 items of 1 unit weight.
As there are repetitions of the same subproblem again and again we can implement the following idea to solve the problem.
If we get a subproblem the first time, we can solve this problem by creating a 2-D array that can store a particular state (n, w). Now if we come across the same state (n, w) again instead of calculating it in exponential complexity we can directly return its result stored in the table in constant time.
Below is the implementation of the above approach:
// Here is the top-down approach of// dynamic programming#include <bits/stdc++.h>using namespace std;// Returns the value of maximum profitint knapSackRec(int W, int wt[], int val[], int index, int** dp){ // base condition if (index < 0) return 0; if (dp[index][W] != -1) return dp[index][W]; if (wt[index] > W) { // Store the value of function call // stack in table before return dp[index][W] = knapSackRec(W, wt, val, index - 1, dp); return dp[index][W]; } else { // Store value in a table before return dp[index][W] = max(val[index] + knapSackRec(W - wt[index], wt, val, index - 1, dp), knapSackRec(W, wt, val, index - 1, dp)); // Return value of table after storing return dp[index][W]; }}int knapSack(int W, int wt[], int val[], int n){ // double pointer to declare the // table dynamically int** dp; dp = new int*[n]; // loop to create the table dynamically for (int i = 0; i < n; i++) dp[i] = new int[W + 1]; // loop to initially filled the // table with -1 for (int i = 0; i < n; i++) for (int j = 0; j < W + 1; j++) dp[i][j] = -1; return knapSackRec(W, wt, val, n - 1, dp);}// Driver Codeint main(){ int profit[] = { 60, 100, 120 }; int weight[] = { 10, 20, 30 }; int W = 50; int n = sizeof(profit) / sizeof(profit[0]); cout << knapSack(W, weight, profit, n); return 0;}
// Here is the top-down approach of// dynamic programmingimport java.io.*;class GFG { // A utility function that returns // maximum of two integers static int max(int a, int b) { return (a > b) ? a : b; } // Returns the value of maximum profit static int knapSackRec(int W, int wt[], int val[], int n, int[][] dp) { // Base condition if (n == 0 || W == 0) return 0; if (dp[n][W] != -1) return dp[n][W]; if (wt[n - 1] > W) // Store the value of function call // stack in table before return return dp[n][W] = knapSackRec(W, wt, val, n - 1, dp); else // Return value of table after storing return dp[n][W] = max((val[n - 1] + knapSackRec(W - wt[n - 1], wt, val, n - 1, dp)), knapSackRec(W, wt, val, n - 1, dp)); } static int knapSack(int W, int wt[], int val[], int N) { // Declare the table dynamically int dp[][] = new int[N + 1][W + 1]; // Loop to initially filled the // table with -1 for (int i = 0; i < N + 1; i++) for (int j = 0; j < W + 1; j++) dp[i][j] = -1; return knapSackRec(W, wt, val, N, dp); } // Driver Code public static void main(String[] args) { int profit[] = { 60, 100, 120 }; int weight[] = { 10, 20, 30 }; int W = 50; int N = profit.length; System.out.println(knapSack(W, weight, profit, N)); }}// This Code is contributed By FARAZ AHMAD
# This is the memoization approach of# 0 / 1 Knapsack in Python in simple# we can say recursion + memoization = DPdef knapsack(wt, val, W, n): # base conditions if n == 0 or W == 0: return 0 if t[n][W] != -1: return t[n][W] # choice diagram code if wt[n-1] <= W: t[n][W] = max( val[n-1] + knapsack( wt, val, W-wt[n-1], n-1), knapsack(wt, val, W, n-1)) return t[n][W] elif wt[n-1] > W: t[n][W] = knapsack(wt, val, W, n-1) return t[n][W]# Driver codeif __name__ == '__main__': profit = [60, 100, 120] weight = [10, 20, 30] W = 50 n = len(profit) # We initialize the matrix with -1 at first. t = [[-1 for i in range(W + 1)] for j in range(n + 1)] print(knapsack(weight, profit, W, n))# This code is contributed by Prosun Kumar Sarkar
// Here is the top-down approach of// dynamic programmingusing System;public class GFG { // A utility function that returns // maximum of two integers static int max(int a, int b) { return (a > b) ? a : b; } // Returns the value of maximum profit static int knapSackRec(int W, int[] wt, int[] val, int n, int[, ] dp) { // Base condition if (n == 0 || W == 0) return 0; if (dp[n, W] != -1) return dp[n, W]; if (wt[n - 1] > W) // Store the value of function call // stack in table before return return dp[n, W] = knapSackRec(W, wt, val, n - 1, dp); else // Return value of table after storing return dp[n, W] = max((val[n - 1] + knapSackRec(W - wt[n - 1], wt, val, n - 1, dp)), knapSackRec(W, wt, val, n - 1, dp)); } static int knapSack(int W, int[] wt, int[] val, int N) { // Declare the table dynamically int[, ] dp = new int[N + 1, W + 1]; // Loop to initially filled the // table with -1 for (int i = 0; i < N + 1; i++) for (int j = 0; j < W + 1; j++) dp[i, j] = -1; return knapSackRec(W, wt, val, N, dp); } // Driver Code static public void Main() { int[] profit = new int[] { 60, 100, 120 }; int[] weight = new int[] { 10, 20, 30 }; int W = 50; int N = profit.Length; Console.WriteLine(knapSack(W, weight, profit, N)); }}// This Code is contributed By Dharanendra L V.
// A utility function that returns// maximum of two integers function max(a, b) { return (a > b) ? a : b; }// Returns the value of maximum profitfunction knapSackRec(W, wt, val, n,dp){ // Base condition if (n == 0 || W == 0) return 0; if (dp[n][W] != -1) return dp[n][W]; if (wt[n - 1] > W) // Store the value of function call // stack in table before return return dp[n][W] = knapSackRec(W, wt, val, n - 1, dp); else // Return value of table after storing return dp[n][W] = max((val[n - 1] + knapSackRec(W - wt[n - 1], wt, val, n - 1, dp)), knapSackRec(W, wt, val, n - 1, dp)); }function knapSack( W, wt,val,N){ // Declare the dp table dynamically // Intializing dp tables(row and cols) with -1 below var dp = new Array(N+1).fill(-1).map(el => new Array(W+1).fill(-1)); return knapSackRec(W, wt, val, N, dp); } var profit= [ 60, 100, 120 ]; var weight = [ 10, 20, 30 ]; var W = 50; var N = profit.length; console.log(knapSack(W, weight, profit, N));// This code is contributed by aksh*tsaxenaa09.
Output
220
Time Complexity: O(N * W).As redundant calculations of states are avoided.
Auxiliary Space: O(N * W) + O(N).The use of a 2D array data structure for storing intermediate states and O(N) auxiliary stack space(ASS) has been used for recursion stack
Bottom-up Approach for 0/1 Knapsack Problem:
To solve the problem follow the below idea:
Since subproblems are evaluated again, this problem has Overlapping Sub-problems property. So the 0/1 Knapsack problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, re-computation of the same subproblems can be avoided by constructing a temporary array K[][] in a bottom-up manner.
Illustration:
Below is the illustration of the above approach:
Let, weight[] = {1, 2, 3}, profit[] = {10, 15, 40}, Capacity = 6
- If no element is filled, then the possible profit is 0.
weight⇢
item⇣/0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 2 3
- For filling the first item in the bag: If we follow the above mentioned procedure, the table will look like the following.
weight⇢
item⇣/0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 10 10 10 10 10 10 2 3
- For filling the second item:
When jthWeight = 2, then maximum possible profit is max (10, DP[1][2-2] + 15) = max(10, 15) = 15.
When jthWeight = 3, then maximum possible profit is max(2 not put, 2 is put into bag) = max(DP[1][3], 15+DP[1][3-2]) = max(10, 25) = 25.
weight⇢
item⇣/0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 10 10 10 10 10 10 2 0 10 15 25 25 25 25 3
- For filling the third item:
When jthWeight = 3, the maximum possible profit is max(DP[2][3], 40+DP[2][3-3]) = max(25, 40) = 40.
When jthWeight = 4, the maximum possible profit is max(DP[2][4], 40+DP[2][4-3]) = max(25, 50) = 50.
When jthWeight = 5, the maximum possible profit is max(DP[2][5], 40+DP[2][5-3]) = max(25, 55) = 55.
When jthWeight = 6, the maximum possible profit is max(DP[2][6], 40+DP[2][6-3]) = max(25, 65) = 65.
weight⇢
item⇣/0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 10 10 10 10 10 10 2 0 10 15 25 25 25 25 3 0 10 15 40 50 55 65
Below is the implementation of the above approach:
// A dynamic programming based// solution for 0-1 Knapsack problem#include <bits/stdc++.h>using namespace std;// A utility function that returns// maximum of two integersint max(int a, int b) { return (a > b) ? a : b; }// Returns the maximum value that// can be put in a knapsack of capacity Wint knapSack(int W, int wt[], int val[], int n){ int i, w; vector<vector<int> > K(n + 1, vector<int>(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]; } } return K[n][W];}// Driver Codeint main(){ int profit[] = { 60, 100, 120 }; int weight[] = { 10, 20, 30 }; int W = 50; int n = sizeof(profit) / sizeof(profit[0]); cout << knapSack(W, weight, profit, n); return 0;}// This code is contributed by Debojyoti Mandal
// A Dynamic Programming based// solution for 0-1 Knapsack problem#include <stdio.h>// A utility function that returns// maximum of two integersint max(int a, int b) { return (a > b) ? a : b; }// Returns the maximum value that// can be put in a knapsack of capacity Wint knapSack(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]; } } return K[n][W];}// Driver Codeint main(){ int profit[] = { 60, 100, 120 }; int weight[] = { 10, 20, 30 }; int W = 50; int n = sizeof(profit) / sizeof(profit[0]); printf("%d", knapSack(W, weight, profit, n)); return 0;}
// A Dynamic Programming based solution// for 0-1 Knapsack problemimport java.io.*;class Knapsack { // A utility function that returns // maximum of two integers static int max(int a, int b) { return (a > b) ? a : b; } // Returns the maximum value that can // be put in a knapsack of capacity W static int knapSack(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] = max(val[i - 1] + K[i - 1][w - wt[i - 1]], K[i - 1][w]); else K[i][w] = K[i - 1][w]; } } return K[n][W]; } // Driver code public static void main(String args[]) { int profit[] = new int[] { 60, 100, 120 }; int weight[] = new int[] { 10, 20, 30 }; int W = 50; int n = profit.length; System.out.println(knapSack(W, weight, profit, n)); }}/*This code is contributed by Rajat Mishra */
# A Dynamic Programming based Python# Program for 0-1 Knapsack problem# Returns the maximum value that can# be put in a knapsack of capacity Wdef knapSack(W, wt, val, n): K = [[0 for x in range(W + 1)] for x 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] return K[n][W]# Driver codeif __name__ == '__main__': profit = [60, 100, 120] weight = [10, 20, 30] W = 50 n = len(profit) print(knapSack(W, weight, profit, n))# This code is contributed by Bhavya Jain
// A Dynamic Programming based solution for// 0-1 Knapsack problemusing System;class GFG { // A utility function that returns // maximum of two integers static int max(int a, int b) { return (a > b) ? a : b; } // Returns the maximum value that // can be put in a knapsack of // capacity W static int knapSack(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]; } } return K[n, W]; } // Driver code static void Main() { int[] profit = new int[] { 60, 100, 120 }; int[] weight = new int[] { 10, 20, 30 }; int W = 50; int n = profit.Length; Console.WriteLine(knapSack(W, weight, profit, n)); }}// This code is contributed by Sam007
// A Dynamic Programming based solution // for 0-1 Knapsack problem // A utility function that returns // maximum of two integers function max(a, b) { return (a > b) ? a : b; } // Returns the maximum value that can // be put in a knapsack of capacity W function knapSack(W, wt, val, n) { let i, w; let K = new Array(n + 1); // Build table K[][] in bottom up manner for (i = 0; i <= n; i++) { K[i] = new Array(W + 1); 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]; } } return K[n][W]; } let profit = [ 60, 100, 120 ]; let weight = [ 10, 20, 30 ]; let W = 50; let n = profit.length; console.log(knapSack(W, weight, profit, n));
<?php// A Dynamic Programming based solution// for 0-1 Knapsack problem// Returns the maximum value that// can be put in a knapsack of // capacity Wfunction knapSack($W, $wt, $val, $n){ $K = array(array()); // Build table K[][] in // bottom up manner for ($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]; } } return $K[$n][$W];} // Driver Code $profit = array(60, 100, 120); $weight = array(10, 20, 30); $W = 50; $n = count($profit); echo knapSack($W, $weight, $profit, $n); // This code is contributed by Sam007.?>
Output
220
Time Complexity: O(N * W).where ‘N’ is the number of elements and ‘W’ is capacity.
Auxiliary Space: O(N * W).The use of a 2-D array of size ‘N*W’.
Space optimized Approach for 0/1 Knapsack Problem using Dynamic Programming:
To solve the problem follow the below idea:
For calculating the current row of the dp[] array we require only previous row, but if we start traversing the rows from right to left then it can be done with a single row only
Below is the implementation of the above approach:
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the maximum profitint knapSack(int W, int wt[], int val[], int n){ // Making and initializing dp array int dp[W + 1]; memset(dp, 0, sizeof(dp)); for (int i = 1; i < n + 1; i++) { for (int w = W; w >= 0; w--) { if (wt[i - 1] <= w) // Finding the maximum value dp[w] = max(dp[w], dp[w - wt[i - 1]] + val[i - 1]); } } // Returning the maximum value of knapsack return dp[W];}// Driver codeint main(){ int profit[] = { 60, 100, 120 }; int weight[] = { 10, 20, 30 }; int W = 50; int n = sizeof(profit) / sizeof(profit[0]); cout << knapSack(W, weight, profit, n); return 0;}
// Java program for the above approachimport java.util.*;class GFG { static int knapSack(int W, int wt[], int val[], int n) { // Making and initializing dp array int[] dp = new int[W + 1]; for (int i = 1; i < n + 1; i++) { for (int w = W; w >= 0; w--) { if (wt[i - 1] <= w) // Finding the maximum value dp[w] = Math.max(dp[w], dp[w - wt[i - 1]] + val[i - 1]); } } // Returning the maximum value of knapsack return dp[W]; } // Driver code public static void main(String[] args) { int profit[] = { 60, 100, 120 }; int weight[] = { 10, 20, 30 }; int W = 50; int n = profit.length; System.out.print(knapSack(W, weight, profit, n)); }}// This code is contributed by gauravrajput1
# Python code to implement the above approachdef knapSack(W, wt, val, n): # Making the dp array dp = [0 for i in range(W+1)] # Taking first i elements for i in range(1, n+1): # Starting from back, # so that we also have data of # previous computation when taking i-1 items for w in range(W, 0, -1): if wt[i-1] <= w: # Finding the maximum value dp[w] = max(dp[w], dp[w-wt[i-1]]+val[i-1]) # Returning the maximum value of knapsack return dp[W]# Driver codeif __name__ == '__main__': profit = [60, 100, 120] weight = [10, 20, 30] W = 50 n = len(profit) print(knapSack(W, weight, profit, n))# This code is contributed by Suyash Saxena
// Java program for the above approachusing System;public class GFG { static int knapSack(int W, int[] wt, int[] val, int n) { // Making and initializing dp array int[] dp = new int[W + 1]; for (int i = 1; i < n + 1; i++) { for (int w = W; w >= 0; w--) { if (wt[i - 1] <= w) // Finding the maximum value dp[w] = Math.Max(dp[w], dp[w - wt[i - 1]] + val[i - 1]); } } // Returning the maximum value of knapsack return dp[W]; } // Driver code public static void Main(String[] args) { int[] profit = { 60, 100, 120 }; int[] weight = { 10, 20, 30 }; int W = 50; int n = profit.Length; Console.Write(knapSack(W, weight, profit, n)); }}// This code is contributed by gauravrajput1
function knapSack(W , wt , val , n) { // Making and initializing dp array var dp = Array(W + 1).fill(0); for (i = 1; i < n + 1; i++) { for (w = W; w >= 0; w--) { if (wt[i - 1] <= w) // Finding the maximum value dp[w] = Math.max(dp[w], dp[w - wt[i - 1]] + val[i - 1]); } } // Returning the maximum value of knapsack return dp[W]; } // Driver code var profit = [ 60, 100, 120 ]; var weight = [ 10, 20, 30 ]; var W = 50; var n = profit.length; console.log(knapSack(W, weight, profit, n));// This code is contributed by Rajput-Ji
Output
220
Time Complexity: O(N * W).As redundant calculations of states are avoided
Auxiliary Space: O(W) As we are using a 1-D array instead of a 2-D array
