Given an array ‘arr’ containing the weight of ‘N’ distinct items, and two knapsacks that can withstand ‘W1’ and ‘W2’ weights, the task is to find the sum of the largest subset of the array ‘arr’, that can be fit in the two knapsacks. It’s not allowed to break any items in two, i.e an item should be put in one of the bags as a whole.
Examples:
Input : arr[] = {8, 3, 2}
W1 = 10, W2 = 3
Output : 13
First and third objects go in the first knapsack. The second object goes in the second knapsack. Thus, the total weight becomes 13.
Input : arr[] = {8, 5, 3}
W1 = 10, W2 = 3
Output : 11
Solution:
A recursive solution is to try out all the possible ways of filling the two knapsacks and choose the one giving the maximum weight.
To optimize the above idea, we need to determine the states of DP, that we will build up our solution upon. After little observation, we can determine that this can be represented in three states (i, w1_r, w2_r). Here ‘i’ means the index of the element we are trying to store, w1_r means the remaining space of the first knapsack, and w2_r means the remaining space of the second knapsack. Thus, the problem can be solved using a 3-dimensional dynamic-programming with a recurrence relation
DP[i][w1_r][w2_r] = max( DP[i + 1][w1_r][w2_r],
arr[i] + DP[i + 1][w1_r - arr[i]][w2_r],
arr[i] + DP[i + 1][w1_r][w2_r - arr[i]])
The explanation for the above recurrence relation is as follows:
For each ‘i’, we can either:
- Don’t select the item ‘i’.
- Fill the item ‘i’ in first knapsack.
- Fill the item ‘i’ in second knapsack.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>#define maxN 31#define maxW 31using namespace std;// 3D array to store// states of DPint dp[maxN][maxW][maxW];// w1_r represents remaining capacity of 1st knapsack// w2_r represents remaining capacity of 2nd knapsack// i represents index of the array arr we are working onint maxWeight(int* arr, int n, int w1_r, int w2_r, int i){ // Base case if (i == n) return 0; if (dp[i][w1_r][w2_r] != -1) return dp[i][w1_r][w2_r]; // Variables to store the result of three // parts of recurrence relation int fill_w1 = 0, fill_w2 = 0, fill_none = 0; if (w1_r >= arr[i]) fill_w1 = arr[i] + maxWeight(arr, n, w1_r - arr[i], w2_r, i + 1); if (w2_r >= arr[i]) fill_w2 = arr[i] + maxWeight(arr, n, w1_r, w2_r - arr[i], i + 1); fill_none = maxWeight(arr, n, w1_r, w2_r, i + 1); // Store the state in the 3D array dp[i][w1_r][w2_r] = max(fill_none, max(fill_w1, fill_w2)); return dp[i][w1_r][w2_r];}// Driver codeint main(){ // Input array int arr[] = { 8, 2, 3 }; // Initializing the array with -1 memset(dp, -1, sizeof(dp)); // Number of elements in the array int n = sizeof(arr) / sizeof(arr[0]); // Capacity of knapsacks int w1 = 10, w2 = 3; // Function to be called cout << maxWeight(arr, n, w1, w2, 0); return 0;} |
Java
// Java implementation of the above approachclass GFG{ static int maxN = 31; static int maxW = 31; // 3D array to store // states of DP static int dp [][][] = new int[maxN][maxW][maxW]; // w1_r represents remaining capacity of 1st knapsack // w2_r represents remaining capacity of 2nd knapsack // i represents index of the array arr we are working on static int maxWeight(int arr [] , int n, int w1_r, int w2_r, int i) { // Base case if (i == n) return 0; if (dp[i][w1_r][w2_r] != -1) return dp[i][w1_r][w2_r]; // Variables to store the result of three // parts of recurrence relation int fill_w1 = 0, fill_w2 = 0, fill_none = 0; if (w1_r >= arr[i]) fill_w1 = arr[i] + maxWeight(arr, n, w1_r - arr[i], w2_r, i + 1); if (w2_r >= arr[i]) fill_w2 = arr[i] + maxWeight(arr, n, w1_r, w2_r - arr[i], i + 1); fill_none = maxWeight(arr, n, w1_r, w2_r, i + 1); // Store the state in the 3D array dp[i][w1_r][w2_r] = Math.max(fill_none, Math.max(fill_w1, fill_w2)); return dp[i][w1_r][w2_r]; } // Driver code public static void main (String[] args) { // Input array int arr[] = { 8, 2, 3 }; // Initializing the array with -1 for (int i = 0; i < maxN ; i++) for (int j = 0; j < maxW ; j++) for (int k = 0; k < maxW ; k++) dp[i][j][k] = -1; // Number of elements in the array int n = arr.length; // Capacity of knapsacks int w1 = 10, w2 = 3; // Function to be called System.out.println(maxWeight(arr, n, w1, w2, 0)); }}// This code is contributed by ihritik |
Python3
# Python3 implementation of the above approach # w1_r represents remaining capacity of 1st knapsack # w2_r represents remaining capacity of 2nd knapsack # i represents index of the array arr we are working on def maxWeight(arr, n, w1_r, w2_r, i): # Base case if i == n: return 0 if dp[i][w1_r][w2_r] != -1: return dp[i][w1_r][w2_r] # Variables to store the result of three # parts of recurrence relation fill_w1, fill_w2, fill_none = 0, 0, 0 if w1_r >= arr[i]: fill_w1 = arr[i] + maxWeight(arr, n, w1_r - arr[i], w2_r, i + 1) if w2_r >= arr[i]: fill_w2 = arr[i] + maxWeight(arr, n, w1_r, w2_r - arr[i], i + 1) fill_none = maxWeight(arr, n, w1_r, w2_r, i + 1) # Store the state in the 3D array dp[i][w1_r][w2_r] = max(fill_none, max(fill_w1, fill_w2)) return dp[i][w1_r][w2_r] # Driver code if __name__ == "__main__": # Input array arr = [8, 2, 3] maxN, maxW = 31, 31 # 3D array to store # states of DP dp = [[[-1] * maxW] * maxW] * maxN # Number of elements in the array n = len(arr) # Capacity of knapsacks w1, w2 = 10, 3 # Function to be called print(maxWeight(arr, n, w1, w2, 0)) # This code is contributed by Rituraj Jain |
C#
// C# implementation of the above approachusing System;class GFG{ static int maxN = 31; static int maxW = 31; // 3D array to store // states of DP static int [ , , ] dp = new int[maxN, maxW, maxW]; // w1_r represents remaining capacity of 1st knapsack // w2_r represents remaining capacity of 2nd knapsack // i represents index of the array arr we are working on static int maxWeight(int [] arr, int n, int w1_r, int w2_r, int i) { // Base case if (i == n) return 0; if (dp[i ,w1_r, w2_r] != -1) return dp[i, w1_r, w2_r]; // Variables to store the result of three // parts of recurrence relation int fill_w1 = 0, fill_w2 = 0, fill_none = 0; if (w1_r >= arr[i]) fill_w1 = arr[i] + maxWeight(arr, n, w1_r - arr[i], w2_r, i + 1); if (w2_r >= arr[i]) fill_w2 = arr[i] + maxWeight(arr, n, w1_r, w2_r - arr[i], i + 1); fill_none = maxWeight(arr, n, w1_r, w2_r, i + 1); // Store the state in the 3D array dp[i, w1_r, w2_r] = Math.Max(fill_none, Math.Max(fill_w1, fill_w2)); return dp[i, w1_r, w2_r]; } // Driver code public static void Main () { // Input array int [] arr = { 8, 2, 3 }; // Initializing the array with -1 for (int i = 0; i < maxN ; i++) for (int j = 0; j < maxW ; j++) for (int k = 0; k < maxW ; k++) dp[i, j, k] = -1; // Number of elements in the array int n = arr.Length; // Capacity of knapsacks int w1 = 10, w2 = 3; // Function to be called Console.WriteLine(maxWeight(arr, n, w1, w2, 0)); }}// This code is contributed by ihritik |
Javascript
<script>// Javascript implementation of // the above approachvar maxN = 31var maxW = 31// 3D array to store// states of DPvar dp = Array(maxN);for(var i=0;i<maxN;i++){ dp[i] = Array(maxW); for(var j =0; j<maxW; j++) { dp[i][j] = Array(maxW).fill(-1); }}// w1_r represents remaining// capacity of 1st knapsack// w2_r represents remaining// capacity of 2nd knapsack// i represents index of the array arr// we are working onfunction maxWeight(arr, n, w1_r, w2_r, i){ // Base case if (i == n) return 0; if (dp[i][w1_r][w2_r] != -1) return dp[i][w1_r][w2_r]; // Variables to store the result of three // parts of recurrence relation var fill_w1 = 0, fill_w2 = 0, fill_none = 0; if (w1_r >= arr[i]) fill_w1 = arr[i] + maxWeight(arr, n, w1_r - arr[i], w2_r, i + 1); if (w2_r >= arr[i]) fill_w2 = arr[i] + maxWeight(arr, n, w1_r, w2_r - arr[i], i + 1); fill_none = maxWeight(arr, n, w1_r, w2_r, i + 1); // Store the state in the 3D array dp[i][w1_r][w2_r] = Math.max(fill_none, Math.max(fill_w1, fill_w2)); return dp[i][w1_r][w2_r];}// Driver code// Input arrayvar arr = [8, 2, 3 ];// Number of elements in the arrayvar n = arr.length;// Capacity of knapsacksvar w1 = 10, w2 = 3;// Function to be calleddocument.write( maxWeight(arr, n, w1, w2, 0));</script> |
Output
13
Time complexity: O(N*W1*W2)
Auxiliary Space: O(N*W1*W2))
Efficient approach : Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a 3-D DP to store the solution of the subproblems.
- Initialize the DP with base cases if (i == 0) then dp[i][w1_r][w2_r] = 0
- Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
- Return the final solution stored in return dp[n][w1][w2].
Implementation :
C++
// C++ implementation of the tabulation approach#include <bits/stdc++.h>#define maxN 31#define maxW 31using namespace std;// 3D array to store// states of DPint dp[maxN][maxW][maxW];// Function to calculate// maximum weight that can be put// into the knapsacksint maxWeight(int* arr, int n, int w1, int w2){ // Iterate over all possible states for (int i = 0; i <= n; i++) { for (int w1_r = 0; w1_r <= w1; w1_r++) { for (int w2_r = 0; w2_r <= w2; w2_r++) { // Base case if (i == 0) dp[i][w1_r][w2_r] = 0; // If the current item can be put in // the first knapsack else if (arr[i - 1] <= w1_r && arr[i - 1] <= w2_r) dp[i][w1_r][w2_r] = max(arr[i - 1] + dp[i - 1][w1_r - arr[i - 1]][w2_r], max(arr[i - 1] + dp[i - 1][w1_r][w2_r - arr[i - 1]], dp[i - 1][w1_r][w2_r])); // If the current item can be put in // the second knapsack else if (arr[i - 1] <= w2_r) dp[i][w1_r][w2_r] = max(arr[i - 1] + dp[i - 1][w1_r][w2_r - arr[i - 1]], dp[i - 1][w1_r][w2_r]); // If the current item can be put in // the first knapsack else if (arr[i - 1] <= w1_r) dp[i][w1_r][w2_r] = max(arr[i - 1] + dp[i - 1][w1_r - arr[i - 1]][w2_r], dp[i - 1][w1_r][w2_r]); // If the current item can not be put in // either of the knapsacks else dp[i][w1_r][w2_r] = dp[i - 1][w1_r][w2_r]; } } } // Return the maximum weight that can be put // into the knapsacks return dp[n][w1][w2];}// Driver codeint main(){ // Input array int arr[] = { 8, 2, 3 }; // Number of elements in the array int n = sizeof(arr) / sizeof(arr[0]); // Capacity of knapsacks int w1 = 10, w2 = 3; // Function to be called cout << maxWeight(arr, n, w1, w2); return 0;} |
Java
import java.util.*;public class Main { // 3D array to store states of DP static int[][][] dp = new int[31][31][31]; // Function to calculate maximum weight // that can be put into the knapsacks public static int maxWeight(int[] arr, int n, int w1, int w2) { // Iterate over all possible states for (int i = 0; i <= n; i++) { for (int w1_r = 0; w1_r <= w1; w1_r++) { for (int w2_r = 0; w2_r <= w2; w2_r++) { // Base case if (i == 0) { dp[i][w1_r][w2_r] = 0; } // If the current item can be put in the first knapsack else if (arr[i - 1] <= w1_r && arr[i - 1] <= w2_r) { dp[i][w1_r][w2_r] = Math.max( arr[i - 1] + dp[i - 1][w1_r - arr[i - 1]][w2_r], Math.max(arr[i - 1] + dp[i - 1][w1_r][w2_r - arr[i - 1]], dp[i - 1][w1_r][w2_r])); } // If the current item can be put in the second knapsack else if (arr[i - 1] <= w2_r) { dp[i][w1_r][w2_r] = Math.max(arr[i - 1] + dp[i - 1][w1_r][w2_r - arr[i - 1]], dp[i - 1][w1_r][w2_r]); } // If the current item can be put in the first knapsack else if (arr[i - 1] <= w1_r) { dp[i][w1_r][w2_r] = Math.max(arr[i - 1] + dp[i - 1][w1_r - arr[i - 1]][w2_r], dp[i - 1][w1_r][w2_r]); } // If the current item can not be put in either of the knapsacks else { dp[i][w1_r][w2_r] = dp[i - 1][w1_r][w2_r]; } } } } // Return the maximum weight that can be put into the knapsacks return dp[n][w1][w2]; } // Driver code public static void main(String[] args) { // Input array int[] arr = { 8, 2, 3 }; // Number of elements in the array int n = arr.length; // Capacity of knapsacks int w1 = 10, w2 = 3; // Function to be called System.out.println(maxWeight(arr, n, w1, w2)); }} |
Python3
# Constants for maximum valuesmaxN = 31maxW = 31# 3D array to store states of DPdp = [[[0 for _ in range(maxW)] for _ in range(maxW)] for _ in range(maxN)]# Function to calculate maximum weight that can be put into the knapsacksdef maxWeight(arr, n, w1, w2): # Iterate over all possible states for i in range(n + 1): for w1_r in range(w1 + 1): for w2_r in range(w2 + 1): # Base case if i == 0: dp[i][w1_r][w2_r] = 0 # If the current item can be put in both knapsacks elif arr[i - 1] <= w1_r and arr[i - 1] <= w2_r: dp[i][w1_r][w2_r] = max( arr[i - 1] + dp[i - 1][w1_r - arr[i - 1]][w2_r], max( arr[i - 1] + dp[i - 1][w1_r][w2_r - arr[i - 1]], dp[i - 1][w1_r][w2_r] ) ) # If the current item can be put in the second knapsack elif arr[i - 1] <= w2_r: dp[i][w1_r][w2_r] = max( arr[i - 1] + dp[i - 1][w1_r][w2_r - arr[i - 1]], dp[i - 1][w1_r][w2_r] ) # If the current item can be put in the first knapsack elif arr[i - 1] <= w1_r: dp[i][w1_r][w2_r] = max( arr[i - 1] + dp[i - 1][w1_r - arr[i - 1]][w2_r], dp[i - 1][w1_r][w2_r] ) # If the current item cannot be put in either knapsack else: dp[i][w1_r][w2_r] = dp[i - 1][w1_r][w2_r] # Return the maximum weight that can be put into the knapsacks return dp[n][w1][w2]# Driver codeif __name__ == "__main__": # Input array arr = [8, 2, 3] # Number of elements in the array n = len(arr) # Capacity of knapsacks w1 = 10 w2 = 3 # Function call print(maxWeight(arr, n, w1, w2)) |
C#
using System;public class KnapsackProblem{ // Function to calculate // maximum weight that can be put // into the knapsacks public static int MaxWeight(int[] arr, int n, int w1, int w2) { // 3D array to store // states of DP int[,,] dp = new int[n + 1, w1 + 1, w2 + 1]; // Iterate over all possible states for (int i = 0; i <= n; i++) { for (int w1_r = 0; w1_r <= w1; w1_r++) { for (int w2_r = 0; w2_r <= w2; w2_r++) { // Base case if (i == 0) dp[i, w1_r, w2_r] = 0; // If the current item can be put in // the first knapsack else if (arr[i - 1] <= w1_r && arr[i - 1] <= w2_r) dp[i, w1_r, w2_r] = Math.Max(arr[i - 1] + dp[i - 1, w1_r - arr[i - 1], w2_r], Math.Max(arr[i - 1] + dp[i - 1, w1_r, w2_r - arr[i - 1]], dp[i - 1, w1_r, w2_r])); // If the current item can be put in // the second knapsack else if (arr[i - 1] <= w2_r) dp[i, w1_r, w2_r] = Math.Max(arr[i - 1] + dp[i - 1, w1_r, w2_r - arr[i - 1]], dp[i - 1, w1_r, w2_r]); // If the current item can be put in // the first knapsack else if (arr[i - 1] <= w1_r) dp[i, w1_r, w2_r] = Math.Max(arr[i - 1] + dp[i - 1, w1_r - arr[i - 1], w2_r], dp[i - 1, w1_r, w2_r]); // If the current item can not be put in // either of the knapsacks else dp[i, w1_r, w2_r] = dp[i - 1, w1_r, w2_r]; } } } // Return the maximum weight that can be put // into the knapsacks return dp[n, w1, w2]; } // Driver code public static void Main() { // Input array int[] arr = { 8, 2, 3 }; // Number of elements in the array int n = arr.Length; // Capacity of knapsacks int w1 = 10, w2 = 3; // Function to be called Console.WriteLine(MaxWeight(arr, n, w1, w2)); }} |
Javascript
const maxN = 31;const maxW = 31;// 3D array to store states of DPconst dp = new Array(maxN).fill(null).map(() => new Array(maxW).fill(null).map(() => new Array(maxW).fill(0)));// Function to calculate maximum weight that can be put into the knapsacksfunction maxWeight(arr, n, w1, w2) { // Iterate over all possible states for (let i = 0; i <= n; i++) { for (let w1_r = 0; w1_r <= w1; w1_r++) { for (let w2_r = 0; w2_r <= w2; w2_r++) { // Base case if (i === 0) { dp[i][w1_r][w2_r] = 0; } else if (arr[i - 1] <= w1_r && arr[i - 1] <= w2_r) { dp[i][w1_r][w2_r] = Math.max( arr[i - 1] + dp[i - 1][w1_r - arr[i - 1]][w2_r], Math.max( arr[i - 1] + dp[i - 1][w1_r][w2_r - arr[i - 1]], dp[i - 1][w1_r][w2_r] ) ); } else if (arr[i - 1] <= w2_r) { dp[i][w1_r][w2_r] = Math.max( arr[i - 1] + dp[i - 1][w1_r][w2_r - arr[i - 1]], dp[i - 1][w1_r][w2_r] ); } else if (arr[i - 1] <= w1_r) { dp[i][w1_r][w2_r] = Math.max( arr[i - 1] + dp[i - 1][w1_r - arr[i - 1]][w2_r], dp[i - 1][w1_r][w2_r] ); } else { dp[i][w1_r][w2_r] = dp[i - 1][w1_r][w2_r]; } } } } // Return the maximum weight that can be put into the knapsacks return dp[n][w1][w2];}// Driver code// Input arrayconst arr = [8, 2, 3]; // Number of elements in the arrayconst n = arr.length;// Capacity of knapsacksconst w1 = 10, w2 = 3;// Function to be calledconsole.log(maxWeight(arr, n, w1, w2)); |
Output
13
Time complexity: O(N*W1*W2)
Auxiliary Space: O(N*W1*W2))
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