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Double Knapsack | Dynamic Programming
  • Difficulty Level : Hard
  • Last Updated : 12 May, 2021

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: 
 

  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++




// 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;
}

Java




// 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

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 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

Javascript




<script>
 
// Javascript implementation of
// the above approach
 
var maxN = 31
var maxW = 31
 
// 3D array to store
// states of DP
var 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 on
function 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 array
var arr = [8, 2, 3 ];
 
// Number of elements in the array
var n = arr.length;
 
// Capacity of knapsacks
var w1 = 10, w2 = 3;
 
// Function to be called
document.write( maxWeight(arr, n, w1, w2, 0));
 
</script>
Output: 
13

 

Time complexity: O(N*W1*W2).
 

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