Given N items, each item having a given weight Ci and a profit value Pi, the task is to maximize the profit by selecting a maximum of K items adding up to a maximum weight W.
Examples:
Input: N = 5, P[] = {2, 7, 1, 5, 3}, C[] = {2, 5, 2, 3, 4}, W = 8, K = 2.
Output: 12
Explanation:
Here, the maximum possible profit is when we take 2 items: item2 (P[1] = 7 and C[1] = 5) and item4 (P[3] = 5 and C[3] = 3).
Hence, maximum profit = 7 + 5 = 12Input: N = 5, P[] = {2, 7, 1, 5, 3}, C[] = {2, 5, 2, 3, 4}, W = 1, K = 2
Output: 0
Explanation: All weights are greater than 1. Hence, no item can be picked.
Approach: The dynamic programming approach is preferred over the general recursion approach. Let us first verify that the conditions of DP are still satisfied.
- Overlapping sub-problems: When the recursive solution is tried, 1 item is added first and the solution set is (1), (2), …(n). In the second iteration we have (1, 2) and so on where (1) and (2) are recalculated. Hence there will be overlapping solutions.
- Optimal substructure: Overall, each item has only two choices, either it can be included in the solution or denied. For a particular subset of z elements, the solution for (z+1)th element can either have a solution corresponding to the z elements or the (z+1)th element can be added if it doesn’t exceed the knapsack constraints. Either way, the optimal substructure property is satisfied.
Let’s derive the recurrence. Let us consider a 3-dimensional table dp[N][W][K], where N is the number of elements, W is the maximum weight capacity and K is the maximum number of items allowed in the knapsack. Let’s define a state dp[i][j][k] where i denotes that we are considering the ith element, j denotes the current weight filled, and k denotes the number of items filled until now.
For every state dp[i][j][k], the profit is either that of the previous state (when the current state is not included) or the profit of the current item added to that of the previous state (when the current item is selected). Hence, the recurrence relation is:
dp[i][j][k] = max( dp[i-1][j][k], dp[i-1][j-W[i]][k-1] + P[i])
Note: In code we have used 1 based indexing so, we are doing i – 1 for weight and profit array.
Below is the implementation of the above approach:
C++
// C++ code for the extended// Knapsack Approach#include <bits/stdc++.h>using namespace std;// To store the dp valuesvector<vector<vector<int> > > dp;int maxProfit(int profit[], int weight[], int n, int max_W, int max_E){ // for each element given for (int i = 1; i <= n; i++) { // For each possible // weight value for (int j = 1; j <= max_W; j++) { // For each case where // the total elements are // less than the constraint for (int k = 1; k <= max_E; k++) { // To ensure that we dont // go out of the array if (j >= weight[i - 1]) { dp[i][j][k] = max( dp[i - 1][j][k], dp[i - 1][j - weight[i - 1]][k - 1] + profit[i - 1]); } else { dp[i][j][k] = dp[i - 1][j][k]; } } } } return dp[n][max_W][max_E];}// Driver Codeint main(){ int n = 5; int profit[] = { 2, 7, 1, 5, 3 }; int weight[] = { 2, 5, 2, 3, 4 }; int max_weight = 8; int max_element = 2; dp = vector<vector<vector<int> > >( n + 1, vector<vector<int> >( max_weight + 1, vector<int>(max_element + 1, 0))); cout << maxProfit(profit, weight, n, max_weight, max_element) << "\n"; return 0;} |
Java
// Java code for the extended// Knapsack Approachimport java.util.*;import java.lang.*;import java.io.*;class GFG{// To store the dp valuesstatic int[][][] dp = new int[100][100][100];static int maxProfit(int profit[], int weight[], int n, int max_W, int max_E){ // for each element given for(int i = 1; i <= n; i++) { // For each possible // weight value for(int j = 1; j <= max_W; j++) { // For each case where // the total elements are // less than the constraint for(int k = 1; k <= max_E; k++) { // To ensure that we dont // go out of the array if (j >= weight[i - 1]) { dp[i][j][k] = Math.max(dp[i - 1][j][k], dp[i - 1][j - weight[i - 1]][k - 1] + profit[i - 1]); } else { dp[i][j][k] = dp[i - 1][j][k]; } } } } return dp[n][max_W][max_E];} // Driver codepublic static void main(String[] args) { int n = 5; int profit[] = { 2, 7, 1, 5, 3 }; int weight[] = { 2, 5, 2, 3, 4 }; int max_weight = 8; int max_element = 2; System.out.println(maxProfit(profit, weight, n, max_weight, max_element)); }}// This code is contributed by offbeat |
Python3
# Python3 code for the extended# Knapsack Approach# To store the dp valuesdp=[]def maxProfit(profit, weight, n, max_W, max_E): # for each element given for i in range(1,n+1) : # For each possible # weight value for j in range(1,max_W+1) : # For each case where # the total elements are # less than the constra for k in range(1, max_E+1) : # To ensure that we dont # go out of the array if (j >= weight[i - 1]) : dp[i][j][k] = max( dp[i - 1][j][k], dp[i - 1][j - weight[i - 1]][k - 1] + profit[i - 1]) else : dp[i][j][k] = dp[i - 1][j][k] return dp[n][max_W][max_E]# Driver Codeif __name__ == '__main__': n = 5 profit = [2, 7, 1, 5, 3 ] weight = [ 2, 5, 2, 3, 4 ] max_weight = 8 max_element = 2 dp = [[[0 for j in range(max_element + 1)]for i in range(max_weight + 1)] for k in range(n+1)] print(maxProfit(profit, weight, n, max_weight, max_element)) |
C#
// C# program to implement above approachusing System;using System.Collections;using System.Collections.Generic;class GFG{ // To store the dp values static int[][][] dp = new int[100][][]; static int maxProfit(int[] profit, int[] weight, int n, int max_W, int max_E) { // for each element given for(int i = 1 ; i <= n ; i++) { // For each possible // weight value for(int j = 1 ; j <= max_W ; j++) { // For each case where // the total elements are // less than the constraint for(int k = 1 ; k <= max_E ; k++) { // To ensure that we dont // go out of the array if (j >= weight[i - 1]) { dp[i][j][k] = Math.Max(dp[i - 1][j][k], dp[i - 1][j - weight[i - 1]][k - 1] + profit[i - 1]); } else { dp[i][j][k] = dp[i - 1][j][k]; } } } } return dp[n][max_W][max_E]; } // Driver code public static void Main(string[] args){ int n = 5; int[] profit = new int[]{ 2, 7, 1, 5, 3 }; int[] weight = new int[]{ 2, 5, 2, 3, 4 }; int max_weight = 8; int max_element = 2; // Initialize dp for(int i = 0 ; i < 100 ; i++){ dp[i] = new int[100][]; for(int j = 0 ; j < 100 ; j++){ dp[i][j] = new int[100]; } } Console.WriteLine(maxProfit(profit, weight, n, max_weight, max_element)); }}// This code is contributed by subhamgoyal2014. |
Javascript
<script>// Javascript code for the extended// Knapsack Approach// To store the dp valuesvar dp = Array.from(Array(100), ()=>Array(100));for(var i =0; i<100; i++) for(var j =0; j<100; j++) dp[i][j] = new Array(100).fill(0);function maxProfit(profit,weight, n, max_W, max_E){ // for each element given for (var i = 1; i <= n; i++) { // For each possible // weight value for (var j = 1; j <= max_W; j++) { // For each case where // the total elements are // less than the constraint for (var k = 1; k <= max_E; k++) { // To ensure that we dont // go out of the array if (j >= weight[i-1]) { dp[i][j][k] = Math.max(dp[i - 1][j][k], dp[i - 1][j - weight[i-1]][k - 1]+ profit[i-1]); } else { dp[i][j][k] = dp[i - 1][j][k]; } } } } return dp[n][max_W][max_E];}// Driver Codevar n = 5;var profit = [2, 7, 1, 5, 3 ];var weight = [2, 5, 2, 3, 4 ];var max_weight = 8;var max_element = 2;document.write( maxProfit(profit, weight, n, max_weight, max_element) + "<br>");</script> |
Output
12
Time Complexity: O(N * W * K)
Auxiliary Space: O(N * W * K)
Approach: 2 The above code takes O(N * W * K) extra space, however, this can be reduced to a 2d matrix. Note that only space would be reduced to O(K * W), and time complexity remains the same, i.e. O(N * W * K).
- The idea is to eliminate the array storage for items. If you see, we update a cell based on dp[i][j][k] = max( dp[i – 1][j][k], dp[i – 1][j – weight[i ]][k – 1] + profit[i]);
instead, we can only use a 2d matrix where the x-axis would represent knapsack capacity (0, 1, 2, 3…W) and y-axis represents k items picked (0, 1, 2, 3, … k). - Note that a cell is updated keeping only two values in mind, the previous value dp[i – 1][j][k], and the value for remaining capacity with k – 1 items. So we only need the previous row to check for [i – 1 ] and [i – 1][j – weight[i ]].
- A cell dp[j][k] in the matrix represents the maximum profit we can get with capacity j, if at most k, items are allowed, with all the items till i. This way we can still iterate over items but we don’t need to keep storage for them.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>#include <iostream>using namespace std;int maxProfit(int profit[], int weight[], int n, int maxCapacity, int maxItems){ vector<vector<int> > dp( maxItems + 1, vector<int>(maxCapacity + 1, 0)); for (int i = 0; i < n; i++) { for (int k = 1; k < maxItems + 1; k++) { for (int j = maxCapacity; j >= 0; j--) { if (weight[i] <= j) { dp[k][j] = max( dp[k][j], profit[i] + dp[k - 1][j - weight[i]]); } } } } return dp[maxItems][maxCapacity];}// Drivers Codeint main(){ int n = 5; int profit[] = { 2, 7, 1, 5, 3 }; int weight[] = { 2, 5, 2, 3, 4 }; int max_weight = 8; int max_element = 2; // Function Call cout << maxProfit(profit, weight, n, max_weight, max_element) << "\n"; return 0;} |
Java
public class Main { static int maxProfit(int profit[], int weight[], int n, int maxCapacity, int maxItems) { int[][] dp = new int[maxItems + 1][maxCapacity + 1]; for (int i = 0; i < n; i++) { for (int k = 1; k < maxItems + 1; k++) { for (int j = maxCapacity; j >= 0; j--) { if (weight[i] <= j) { dp[k][j] = Math.max( dp[k][j], profit[i] + dp[k - 1][j - weight[i]]); } } } } return dp[maxItems][maxCapacity]; } public static void main(String[] args) { int n = 5; int[] profit = { 2, 7, 1, 5, 3 }; int[] weight = { 2, 5, 2, 3, 4 }; int max_weight = 8; int max_element = 2; System.out.println(maxProfit( profit, weight, n, max_weight, max_element)); }} // this code is contributed by devendra1 |
Python3
def maxProfit(profit, weight, n, maxCapacity, maxItems): dp = [[0 for j in range(maxCapacity + 1)] for i in range(maxItems + 1)] for i in range(n): for k in range(1, maxItems + 1): for j in range(maxCapacity, -1, -1): if weight[i] <= j: dp[k][j] = max(dp[k][j], profit[i] + dp[k - 1][j - weight[i]]) return dp[maxItems][maxCapacity]# Drivers Codeif __name__ == "__main__": n = 5 profit = [2, 7, 1, 5, 3] weight = [2, 5, 2, 3, 4] max_weight = 8 max_element = 2 # Function Call print(maxProfit(profit, weight, n, max_weight, max_element)) |
C#
using System;class GFG { // This function calculates the maximum profit static int MaxProfit(int[] profit, int[] weight, int n, int maxCapacity, int maxItems) { // 2D Array to store DP int[, ] dp = new int[maxItems + 1, maxCapacity + 1]; // Building the DP array for (int i = 0; i < n; i++) { for (int k = 1; k < maxItems + 1; k++) { for (int j = maxCapacity; j >= 0; j--) { if (weight[i] <= j) { dp[k, j] = Math.Max( dp[k, j], profit[i] + dp[k - 1, j - weight[i]]); } } } } return dp[maxItems, maxCapacity]; } // Driver code public static void Main(string[] args) { int n = 5; int[] profit = { 2, 7, 1, 5, 3 }; int[] weight = { 2, 5, 2, 3, 4 }; int max_weight = 8; int max_element = 2; // Function call Console.WriteLine(MaxProfit( profit, weight, n, max_weight, max_element)); }} // this code is contributed by phasing17 |
Javascript
function maxProfit(profit, weight, n, maxCapacity, maxItems) { // initialize a 2D array with zeros for dynamic programming const dp = Array.from(Array(maxItems + 1), () => new Array(maxCapacity + 1).fill(0)); for (let i = 0; i < n; i++) { for (let k = 1; k < maxItems + 1; k++) { for (let j = maxCapacity; j >= 0; j--) { // check if the weight of the item is less than or equal // to the remaining capacity if (weight[i] <= j) { // take the maximum between not including the // current item and including the current item dp[k][j] = Math.max( dp[k][j], // not including the current item profit[i] + dp[k - 1][j - weight[i]] // including the current item ); } } } } return dp[maxItems][maxCapacity]; // return the maximum profit achievable}// example usageconst n = 5;const profit = [2, 7, 1, 5, 3];const weight = [2, 5, 2, 3, 4];const max_weight = 8;const max_element = 2;console.log(maxProfit(profit, weight, n, max_weight, max_element)); |
Output
12
Time Complexity: O(N * W * K)
Auxiliary Space: O(W * K)
Efficient approach : Space optimization
In previous approach the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.
Implementation steps:
- Create a 1D vector dp of size maxCapacity+1.
- Set a base case by initializing the values of DP .
- Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
- At last return and print the final answer stored in dp[maxCapacity].
Implementation:
C++
#include <bits/stdc++.h>#include <iostream>using namespace std;int maxProfit(int profit[], int weight[], int n, int maxCapacity, int maxItems){ vector<int> dp(maxCapacity + 1, 0); for (int i = 0; i < n; i++) { for (int j = maxCapacity; j >= weight[i]; j--) { dp[j] = max(dp[j], profit[i] + dp[j - weight[i]]); } } return dp[maxCapacity];}// Drivers Codeint main(){ int n = 5; int profit[] = { 2, 7, 1, 5, 3 }; int weight[] = { 2, 5, 2, 3, 4 }; int max_weight = 8; int max_element = 2; // Function Call cout << maxProfit(profit, weight, n, max_weight, max_element) << "\n"; return 0;} |
Java
import java.util.Arrays;class GFG { public static int maxProfit(int[] profit, int[] weight, int n, int maxCapacity, int maxItems) { int[] dp = new int[maxCapacity + 1]; Arrays.fill(dp, 0); for (int i = 0; i < n; i++) { for (int j = maxCapacity; j >= weight[i]; j--) { dp[j] = Math.max(dp[j], profit[i] + dp[j - weight[i]]); } } return dp[maxCapacity]; } public static void main(String[] args) { int n = 5; int[] profit = {2, 7, 1, 5, 3}; int[] weight = {2, 5, 2, 3, 4}; int max_weight = 8; int max_element = 2; // Function Call System.out.println(maxProfit(profit, weight, n, max_weight, max_element)); }} |
Python3
def maxProfit(profit, weight, n, maxCapacity, maxItems): dp = [0]*(maxCapacity + 1) for i in range(n): for j in range(maxCapacity, weight[i] - 1, -1): dp[j] = max(dp[j], profit[i] + dp[j - weight[i]]) return dp[maxCapacity]# Drivers Codeif __name__ == '__main__': n = 5 profit = [2, 7, 1, 5, 3] weight = [2, 5, 2, 3, 4] max_weight = 8 max_element = 2 # Function Call print(maxProfit(profit, weight, n, max_weight, max_element)) |
C#
using System;public class GFG{ public static int MaxProfit(int[] profit, int[] weight, int n, int maxCapacity, int maxItems) { int[] dp = new int[maxCapacity + 1]; for (int i = 0; i < n; i++) { for (int j = maxCapacity; j >= weight[i]; j--) { dp[j] = Math.Max(dp[j], profit[i] + dp[j - weight[i]]); } } return dp[maxCapacity]; } public static void Main(string[] args) { int n = 5; int[] profit = { 2, 7, 1, 5, 3 }; int[] weight = { 2, 5, 2, 3, 4 }; int maxWeight = 8; int maxElement = 2; // Function Call Console.WriteLine(MaxProfit(profit, weight, n, maxWeight, maxElement)); }} |
Javascript
// Function to calculate the maximum profit for a given knapsack problemfunction maxProfit(profit, weight, n, maxCapacity, maxItems) { // Create a dp array to store the maximum profit for each capacity (from 0 to maxCapacity) const dp = new Array(maxCapacity + 1).fill(0); // Loop through each item and calculate the maximum profit for each capacity // using bottom-up approach for (let i = 0; i < n; i++) { for (let j = maxCapacity; j >= weight[i]; j--) { // The maximum profit at capacity 'j' is either the // current profit plus the profit for the remaining capacity (j - weight[i]), // or the previous maximum profit at capacity 'j' dp[j] = Math.max(dp[j], profit[i] + dp[j - weight[i]]); } } // The final value in dp array at maxCapacity represents the maximum // profit that can be obtained // using the given knapsack capacity and number of items constraint return dp[maxCapacity];}// Drivers Code const n = 5; const profit = [2, 7, 1, 5, 3]; const weight = [2, 5, 2, 3, 4]; const maxWeight = 8; const maxElement = 2; // Function Call console.log(maxProfit(profit, weight, n, maxWeight, maxElement)); |
Output
12
Time Complexity: O(N * W * K)
Auxiliary Space: O(K)
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