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. 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:
 

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 :

13

Time complexity: O(N*W1*W2)
Auxiliary Space: O(N*W1*W2))