Double Knapsack | Dynamic Programming

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. Its 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 first knapsack, and w2_r means the remaining space of 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:

  1. Don’t select the item ‘i’.
  2. Fill the item ‘i’ in first knapsack.
  3. Fill the item ‘i’ in second knapsack.

Below is the implementation of the above approach:

C++

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// C++ implementation of the above approach
#include <bits/stdc++.h>
#define maxN 31
#define maxW 31
using namespace std;
  
// 3D array to store
// states of DP
int 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 on
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] = max(fill_none, max(fill_w1, fill_w2));
  
    return dp[i][w1_r][w2_r];
}
  
// Driver code
int 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;
}

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Java














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// Java implementation of the above approach 


  


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 (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 



















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Python3














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# 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 



















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C#














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// C# implementation of the above approach 


using 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 



















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Output:


13







Time complexity: O(N*W1*W2).


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