Double Knapsack | Dynamic Programming
Last Updated :
11 Jul, 2025
Given an array arr[] containing the weight of 'n' distinct items, and two knapsacks that can withstand capactiy1 and capacity2 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], capacity1 = 10, capacity2 = 3
Output: 13
Explanation: The first and third items are placed in the first knapsack, while the second item is placed in the second knapsack. This results in a total weight of 13.
Input: arr[] = [8, 5, 3] capacity1 = 10, capacity2 = 3
Output: 11
Explanation: The first item is placed in the first knapsack, and the third item is placed in the second knapsack. This results in a total weight of 11.
Using Recursion– O(2^n) Time and O(n) Space
A recursive solutions is to try out all the possible ways of filling the two knapsacks and choose the one giving the maximum weight.
Every item has 3 choices:
Don't include the current item: If we don't include the current item (i.e., arr[curr]), the maximum weight is simply the same as if we move on to the next item, keeping both knapsack capacities unchanged:
- findMaxWeight(curr, n, arr, capacity1, capacity2) = findMaxWeight(curr+1, n, arr, capacity1, capacity2)
Include the current item in the first knapsack: If the current item can fit into the first knapsack (i.e., arr[curr] ≤ capacity1), then we include it in the first knapsack, reduce the capacity of the first knapsack by arr[curr], and proceed to the next item with updated capacities. The recurrence relation for this choice is:
- findMaxWeight(curr, n, arr, capacity1, capacity2) = max(findMaxWeight(curr+1, n, arr, capacity1-arr[cur], capacity2)+arr[curr], findMaxWeight(curr+1, n, arr, capacity1, capacity2))
Include the current item in the second knapsack: If the current item can fit into the second knapsack (i.e., arr[curr] ≤ capacity2), then we include it in the second knapsack, reduce the capacity of the second knapsack by arr[curr], and proceed to the next item with updated capacities. The recurrence relation for this choice is:
- findMaxWeight(curr, n, arr, capacity1, capacity2) = max(findMaxWeight(curr+1, n, arr, capacity1, capacity2-arr[curr])+arr[curr], findMaxWeight(curr+1, n, arr, capacity1, capacity2))
C++
// C++ program to find the largest subset of
// the array that can fit into two knapsack
// using recursion
#include <iostream>
#include <vector>
using namespace std;
int findMaxWeight(int curr, int n, vector<int> &arr,
int capacity1, int capacity2) {
// Base case: if all items have been considered
if (curr >= n)
return 0;
// Option 1: Don't take the current item
int res = findMaxWeight(curr + 1, n, arr,
capacity1, capacity2);
// Option 2: If the current item can be
// added to the first knapsack, do it
if (capacity1 >= arr[curr]) {
int takeInFirst = arr[curr] +
findMaxWeight(curr + 1, n, arr,
capacity1 - arr[curr], capacity2);
res = max(res, takeInFirst);
}
// Option 3: If the current item can be
// added to the second knapsack, do it
if (capacity2 >= arr[curr]) {
int takeInSecond = arr[curr] +
findMaxWeight(curr + 1, n, arr,
capacity1, capacity2 - arr[curr]);
res = max(res, takeInSecond);
}
return res;
}
int maxWeight(vector<int> &arr, int capacity1, int capacity2) {
int n = arr.size();
int res = findMaxWeight(0, n, arr, capacity1, capacity2);
return res;
}
int main() {
vector<int> arr = {8, 2, 3};
int capacity1 = 10, capacity2 = 3;
int res = maxWeight(arr, capacity1, capacity2);
cout << res;
return 0;
}
// Java program to find the largest subset of
// the array that can fit into two knapsacks
// using recursion
import java.util.*;
class GfG {
static int findMaxWeight(int curr, int n, int[] arr,
int capacity1, int capacity2) {
// Base case: if all items have been considered
if (curr >= n)
return 0;
// Option 1: Don't take the current item
int res = findMaxWeight(curr + 1, n, arr,
capacity1, capacity2);
// Option 2: If the current item can be
// added to the first knapsack, do it
if (capacity1 >= arr[curr]) {
int takeInFirst
= arr[curr]
+ findMaxWeight(curr + 1, n, arr,
capacity1 - arr[curr], capacity2);
res = Math.max(res, takeInFirst);
}
// Option 3: If the current item can be
// added to the second knapsack, do it
if (capacity2 >= arr[curr]) {
int takeInSecond
= arr[curr]
+ findMaxWeight(curr + 1, n, arr, capacity1,
capacity2 - arr[curr]);
res = Math.max(res, takeInSecond);
}
return res;
}
static int maxWeight(int[] arr, int capacity1, int capacity2) {
int n = arr.length;
int res = findMaxWeight(0, n, arr, capacity1, capacity2);
return res;
}
public static void main(String[] args) {
int[] arr = { 8, 2, 3 };
int capacity1 = 10, capacity2 = 3;
int res = maxWeight(arr, capacity1, capacity2);
System.out.println(res);
}
}
# Python program to find the largest subset of
# the array that can fit into two knapsacks
# using recursion
def findMaxWeight(curr, n, arr, capacity1, capacity2):
# Base case: if all items have been considered
if curr >= n:
return 0
# Option 1: Don't take the current item
res = findMaxWeight(curr + 1, n, arr, capacity1, capacity2)
# Option 2: If the current item can be
# added to the first knapsack, do it
if capacity1 >= arr[curr]:
takeInFirst = arr[curr] + \
findMaxWeight(curr + 1, n, arr, capacity1 - arr[curr], capacity2)
res = max(res, takeInFirst)
# Option 3: If the current item can be
# added to the second knapsack, do it
if capacity2 >= arr[curr]:
takeInSecond = arr[curr] + \
findMaxWeight(curr + 1, n, arr, capacity1, capacity2 - arr[curr])
res = max(res, takeInSecond)
return res
def maxWeight(arr, capacity1, capacity2):
n = len(arr)
res = findMaxWeight(0, n, arr, capacity1, capacity2)
return res
if __name__ == "__main__":
arr = [8, 2, 3]
capacity1 = 10
capacity2 = 3
res = maxWeight(arr, capacity1, capacity2)
print(res)
// C# program to find the largest subset of
// the array that can fit into two knapsacks
// using recursion
using System;
class GfG {
static int findMaxWeight(int curr, int n, int[] arr,
int capacity1, int capacity2) {
// Base case: if all items have been considered
if (curr >= n)
return 0;
// Option 1: Don't take the current item
int res = findMaxWeight(curr + 1, n, arr,
capacity1, capacity2);
// Option 2: If the current item can be
// added to the first knapsack, do it
if (capacity1 >= arr[curr]) {
int takeInFirst
= arr[curr]
+ findMaxWeight(curr + 1, n, arr,
capacity1 - arr[curr], capacity2);
res = Math.Max(res, takeInFirst);
}
// Option 3: If the current item can be
// added to the second knapsack, do it
if (capacity2 >= arr[curr]) {
int takeInSecond
= arr[curr]
+ findMaxWeight(curr + 1, n, arr, capacity1,
capacity2 - arr[curr]);
res = Math.Max(res, takeInSecond);
}
return res;
}
static int maxWeight(int[] arr, int capacity1, int capacity2) {
int n = arr.Length;
int res = findMaxWeight(0, n, arr, capacity1, capacity2);
return res;
}
static void Main(string[] args) {
int[] arr
= { 8, 2, 3 };
int capacity1 = 10, capacity2
= 3;
int res=maxWeight(arr, capacity1, capacity2);
Console.WriteLine(res);
}
}
// JavaScript program to find the largest subset of
// the array that can fit into two knapsacks
// using recursion
function findMaxWeight(curr, n, arr, capacity1, capacity2) {
// Base case: if all items have been considered
if (curr >= n)
return 0;
// Option 1: Don't take the current item
let res = findMaxWeight(curr + 1, n, arr,
capacity1, capacity2);
// Option 2: If the current item can be
// added to the first knapsack, do it
if (capacity1 >= arr[curr]) {
let takeInFirst
= arr[curr]
+ findMaxWeight(curr + 1, n, arr,
capacity1 - arr[curr], capacity2);
res = Math.max(res, takeInFirst);
}
// Option 3: If the current item can be
// added to the second knapsack, do it
if (capacity2 >= arr[curr]) {
let takeInSecond
= arr[curr]
+ findMaxWeight(curr + 1, n, arr, capacity1,
capacity2 - arr[curr]);
res = Math.max(res, takeInSecond);
}
return res;
}
function maxWeight(arr, capacity1, capacity2) {
let n = arr.length;
let res = findMaxWeight(0, n, arr, capacity1, capacity2);
return res;
}
let arr = [ 8, 2, 3 ];
let capacity1 = 10, capacity2 = 3;
let res = maxWeight(arr, capacity1,capacity2);
console.log(res);
Using Memoization – O(n*capacity1*capacity2) Time and O(n*capacity1*capacity2) Space
If we notice carefully, we can observe that the above recursive solution holds the following two properties of Dynamic Programming.
1. Optimal Substructure: The solution to the two-knapsack problem can be derived from the optimal solutions of smaller subproblems. Specifically, for any given n (the number of items considered) and the remaining capacities of the two knapsacks (capacity1 and capacity2), we can express the recursive relation as follows.
2. Overlapping Subproblems: In the recursive solution to the problem of maximizing the weight in two knapsacks, we observe that many subproblems are computed multiple times, which leads to overlapping subproblems.
- We need to track all three parameters, so we create a 3D memoization table of size (n+1) x (capacity1+1) x (capacity2+1).
- This table is initialized with -1 to indicate that no subproblems have been computed yet.
- If the value at memo[curr][capacity1][capacity2] is not -1, return the already computed result to avoid recomputing the same subproblem.
C++
// C++ program to find the largest subset of
// the array that can fit into two knapsack
// using memoization
#include <iostream>
#include <vector>
using namespace std;
int findMaxWeight(int curr, int n, vector<int> &arr, int capacity1,
int capacity2, vector<vector<vector<int>>> &memo) {
// Base case: If all items have been considered,
// return 0 as no more items can be added
if (curr >= n)
return 0;
// If the result for this subproblem is already
// computed, return the memoized value
if (memo[curr][capacity1][capacity2] != -1)
return memo[curr][capacity1][capacity2];
// Option 1: Don't take the current item, and
// proceed to the next item
int res = findMaxWeight(curr + 1, n, arr,
capacity1, capacity2, memo);
// Option 2: If the current item can be added
// to the first knapsack, include it
if (capacity1 >= arr[curr]) {
int takeInFirst = arr[curr] +
findMaxWeight(curr + 1, n, arr, capacity1 - arr[curr],
capacity2, memo);
res = max(res, takeInFirst);
}
// Option 3: If the current item can be added
// to the second knapsack, include it
if (capacity2 >= arr[curr]) {
int takeInSecond = arr[curr] +
findMaxWeight(curr + 1, n, arr, capacity1,
capacity2 - arr[curr], memo);
res = max(res, takeInSecond);
}
return memo[curr][capacity1][capacity2] = res;
}
int maxWeight(vector<int> &arr, int capacity1, int capacity2) {
int n = arr.size();
vector<vector<vector<int>>> memo =
vector<vector<vector<int>>>(n, vector<vector<int>>
(capacity1 + 1, vector<int>(capacity2 + 1, -1)));
int res = findMaxWeight(0, n, arr, capacity1, capacity2, memo);
return res;
}
int main() {
vector<int> arr = {8, 2, 3};
int capacity1 = 10, capacity2 = 3;
int res=maxWeight(arr, capacity1, capacity2);
cout << res;
return 0;
}
// Java program to find the largest subset of
// the array that can fit into two knapsack
// using memoization
import java.util.*;
class GfG {
static int findMaxWeight(int curr, int n, int[] arr,
int capacity1, int capacity2, int[][][] memo) {
// Base case: if all items have been considered
if (curr >= n) {
return 0;
}
// If the result is already computed for this
// subproblem, return it
if (memo[curr][capacity1][capacity2] != -1) {
return memo[curr][capacity1][capacity2];
}
// Option 1: Don't take the current item
int res
= findMaxWeight(curr + 1, n, arr,
capacity1, capacity2, memo);
// Option 2: If the current item can be added
// to the first knapsack, do it
if (capacity1 >= arr[curr]) {
int takeInFirst
= arr[curr]
+ findMaxWeight(curr + 1, n, arr,
capacity1 - arr[curr], capacity2, memo);
res = Math.max(res, takeInFirst);
}
// Option 3: If the current item can be added to the
// second knapsack, do it
if (capacity2 >= arr[curr]) {
int takeInSecond
= arr[curr]
+ findMaxWeight(curr + 1, n, arr, capacity1,
capacity2 - arr[curr], memo);
res = Math.max(res, takeInSecond);
}
// Store the result in the memoization table and
// return it
memo[curr][capacity1][capacity2] = res;
return res;
}
static int maxWeight(int[] arr, int capacity1, int capacity2) {
int n = arr.length;
// Create a memoization table with all values
// initialized to -1
int[][][] memo = new int[n][capacity1 + 1][capacity2 + 1];
// Initialize memoization table with -1
for (int i = 0; i < n; i++) {
for (int j = 0; j <= capacity1; j++) {
for (int k = 0; k <= capacity2; k++) {
memo[i][j][k] = -1;
}
}
}
return findMaxWeight(0, n, arr, capacity1, capacity2, memo);
}
public static void main(String[] args) {
int[] arr = { 8, 2, 3 };
int capacity1 = 10, capacity2 = 3;
int res = maxWeight(arr, capacity1, capacity2);
System.out.println(res);
}
}
# Python program to find the largest subset of
# the array that can fit into two knapsack
# using memoization
def findMaxWeight(curr, n, arr, capacity1, capacity2, memo):
# Base case: if all items have been considered
if curr >= n:
return 0
# If the result is already computed for
# this subproblem, return it
if memo[curr][capacity1][capacity2] != -1:
return memo[curr][capacity1][capacity2]
# Option 1: Don't take the current item
res = findMaxWeight(curr + 1, n, arr, capacity1, capacity2, memo)
# Option 2: If the current item can be added
# to the first knapsack, do it
if capacity1 >= arr[curr]:
takeInFirst = arr[curr] + \
findMaxWeight(curr + 1, n, arr, capacity1 - arr[curr], capacity2, memo)
res = max(res, takeInFirst)
# Option 3: If the current item can be added
# to the second knapsack, do it
if capacity2 >= arr[curr]:
takeInSecond = arr[curr] + \
findMaxWeight(curr + 1, n, arr, capacity1, capacity2 - arr[curr], memo)
res = max(res, takeInSecond)
# Store the result in the memoization table and
# return it
memo[curr][capacity1][capacity2] = res
return res
def maxWeight(arr, capacity1, capacity2):
n = len(arr)
# Create a memoization table with all values
# initialized to -1
memo = [[[-1 for _ in range(capacity2 + 1)]
for _ in range(capacity1 + 1)] for _ in range(n)]
return findMaxWeight(0, n, arr, capacity1, capacity2, memo)
if __name__ == "__main__":
arr = [8, 2, 3]
capacity1 = 10
capacity2 = 3
res = maxWeight(arr, capacity1, capacity2)
print(res)
// C# program to find the largest subset of
// the array that can fit into two knapsack
// using memoization
using System;
class GfG {
static int findMaxWeight(int curr, int n, int[] arr,
int capacity1, int capacity2, int[, , ] memo) {
// Base case: if all items have been considered
if (curr >= n)
return 0;
// If the result is already computed for this
// subproblem, return it
if (memo[curr, capacity1, capacity2] != -1)
return memo[curr, capacity1, capacity2];
// Option 1: Don't take the current item
int res
= findMaxWeight(curr + 1, n, arr,
capacity1, capacity2, memo);
// Option 2: If the current item can be added to the
// first knapsack, do it
if (capacity1 >= arr[curr]) {
int takeInFirst
= arr[curr]
+ findMaxWeight(curr + 1, n, arr,
capacity1 - arr[curr], capacity2, memo);
res = Math.Max(res, takeInFirst);
}
// Option 3: If the current item can be added to the
// second knapsack, do it
if (capacity2 >= arr[curr]) {
int takeInSecond
= arr[curr]
+ findMaxWeight(curr + 1, n, arr, capacity1,
capacity2 - arr[curr], memo);
res = Math.Max(res, takeInSecond);
}
// Store the result in the memoization table and
// return it
memo[curr, capacity1, capacity2] = res;
return res;
}
static int maxWeight(int[] arr, int capacity1, int capacity2) {
int n = arr.Length;
// Create a memoization table with all values
// initialized to -1
int[, , ] memo = new int[n, capacity1 + 1, capacity2 + 1];
for (int i = 0; i < n; i++) {
for (int j = 0; j <= capacity1; j++) {
for (int k = 0; k <= capacity2; k++) {
memo[i, j, k] = -1;
}
}
}
return findMaxWeight(0, n, arr, capacity1, capacity2, memo);
}
static void Main() {
int[] arr = { 8, 2, 3 };
int capacity1 = 10;
int capacity2 = 3;
int res = maxWeight(arr, capacity1, capacity2);
Console.WriteLine(res);
}
}
// JavaScript program to find the largest subset of
// the array that can fit into two knapsack
// using memoization
function findMaxWeight(curr, n, arr, capacity1, capacity2, memo) {
// Base case: if all items have been considered
if (curr >= n)
return 0;
// If the result is already computed for this
// subproblem, return it
if (memo[curr][capacity1][capacity2] !== -1)
return memo[curr][capacity1][capacity2];
// Option 1: Don't take the current item
let res = findMaxWeight(curr + 1, n, arr, capacity1, capacity2, memo);
// Option 2: If the current item can be added to the
// first knapsack, do it
if (capacity1 >= arr[curr]) {
let takeInFirst
= arr[curr]
+ findMaxWeight(curr + 1, n, arr,
capacity1 - arr[curr], capacity2, memo);
res = Math.max(res, takeInFirst);
}
// Option 3: If the current item can be added to the
// second knapsack, do it
if (capacity2 >= arr[curr]) {
let takeInSecond
= arr[curr]
+ findMaxWeight(curr + 1, n, arr, capacity1,
capacity2 - arr[curr], memo);
res = Math.max(res, takeInSecond);
}
// Store the result in the memoization table and return
// it
memo[curr][capacity1][capacity2] = res;
return res;
}
function maxWeight(arr, capacity1, capacity2) {
let n = arr.length;
// Create a memoization table with all values
// initialized to -1
let memo = new Array(n);
for (let i = 0; i < n; i++) {
memo[i] = new Array(capacity1 + 1);
for (let j = 0; j <= capacity1; j++) {
memo[i][j] = new Array(capacity2 + 1).fill(-1);
}
}
return findMaxWeight(0, n, arr, capacity1, capacity2, memo);
}
let arr = [ 8, 2, 3 ];
let capacity1 = 10;
let capacity2 = 3;
let res = maxWeight(arr, capacity1, capacity2);
console.log(res);
Using Tabulation - O(n*capacity1*capacity2) Time and O(n*capacity1*capacity2) Space
The approach is similar to the previous one. just instead of breaking down the problem recursively, we iteratively build up the solution by calculating in bottom-up manner.
So we will create 3D array dp of size (n + 1) x (capacity1 + 1) x (capacity2 + 1). Here, dp[i][j][k] represents the maximum weight that can be achieved by considering items from arr[0] to arr[i-1] with two knapsacks of remaining capacities j and k.
Base Case:
If no items are considered (i = 0), then no weight can be placed in either knapsack, so:
dp[0][j][k]=0 for all j, k
Recurrence Relation:
For each item i (1<=i<=n), we have three choices:
- Do not include item arr[i-1] in either knapsack
dp[i][j][k] = dp[i-1][j][k]
- Include item arr[i-1] in the first knapsack if its weight arr[i-1] is less than or equal to j:
dp[i][j][k] = max(dp[i][j][k], arr[i-1] + dp[i-1][j - arr[i-1]][k])
- Include item arr[i-1] in the second knapsack if its weight arr[i-1] is less than or equal to k:
dp[i][j][k] = max(dp[i][j][k], arr[i-1] + dp[i-1][j][k - arr[i-1]])
C++
// C++ program to find the largest subset of
// the array that can fit into two knapsack
// using tabulation
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
int maxWeight(vector<int> &arr, int capacity1, int capacity2) {
int n = arr.size();
// Initialize a 3D DP array with dimensions (n+1) x (w1+1) x (w2+1)
vector<vector<vector<int>>> dp(n + 1, vector<vector<int>>
(capacity1 + 1, vector<int>(capacity2 + 1, 0)));
// Fill the DP array iteratively
for (int i = 1; i <= n; ++i) {
int weight = arr[i - 1];
for (int j = 0; j <= capacity1; ++j) {
for (int k = 0; k <= capacity2; ++k) {
// Option 1: Don't take the current item
dp[i][j][k] = dp[i - 1][j][k];
// Option 2: Take the current item in the
// first knapsack, if possible
if (j >= weight) {
dp[i][j][k] = max(dp[i][j][k], weight + dp[i - 1][j - weight][k]);
}
// Option 3: Take the current item in the
// second knapsack, if possible
if (k >= weight) {
dp[i][j][k] = max(dp[i][j][k], weight + dp[i - 1][j][k - weight]);
}
}
}
}
return dp[n][capacity1][capacity2];
}
int main() {
vector<int> arr = {8, 2, 3};
int capacity1 = 10, capacity2 = 3;
int res = maxWeight(arr, capacity1, capacity2);
cout << res;
return 0;
}
// Java program to find the largest subset of
// the array that can fit into two knapsack
// using tabulation
import java.util.Arrays;
import java.util.List;
class GfG {
static int maxWeight(List<Integer> arr, int capacity1,
int capacity2) {
int n = arr.size();
// Initialize a 3D DP array with dimensions (n+1) x
// (w1+1) x (w2+1)
int[][][] dp = new int[n + 1][capacity1 + 1][capacity2 + 1];
// Fill the DP array iteratively
for (int i = 1; i <= n; i++) {
int weight = arr.get(i - 1);
for (int j = 0; j <= capacity1; j++) {
for (int k = 0; k <= capacity2; k++) {
// Option 1: Don't take the current item
dp[i][j][k] = dp[i - 1][j][k];
// Option 2: Take the current item in
// the first knapsack, if possible
if (j >= weight) {
dp[i][j][k] = Math.max(
dp[i][j][k],
weight
+ dp[i - 1][j - weight][k]);
}
// Option 3: Take the current item in
// the second knapsack, if possible
if (k >= weight) {
dp[i][j][k] = Math.max(
dp[i][j][k],
weight
+ dp[i - 1][j][k - weight]);
}
}
}
}
return dp[n][capacity1][capacity2];
}
public static void main(String[] args) {
List<Integer> arr = Arrays.asList(8, 2, 3);
int capacity1 = 10, capacity2 = 3;
int res=maxWeight(arr, capacity1, capacity2);
System.out.println(res);
}
}
# Python program to find the largest subset of
# the array that can fit into two knapsack
# using tabulation
def maxWeight(arr, capacity1, capacity2):
n = len(arr)
# Initialize a 3D DP array with dimensions
# (n+1) x (w1+1) x (w2+1)
dp = [[[0 for _ in range(capacity2 + 1)] for _ in range(capacity1 + 1)]
for _ in range(n + 1)]
# Fill the DP array iteratively
for i in range(1, n + 1):
weight = arr[i - 1]
for j in range(capacity1 + 1):
for k in range(capacity2 + 1):
# Option 1: Don't take the current item
dp[i][j][k] = dp[i - 1][j][k]
# Option 2: Take the current item in the
# first knapsack, if possible
if j >= weight:
dp[i][j][k] = max(dp[i][j][k], weight +
dp[i - 1][j - weight][k])
# Option 3: Take the current item in the
# second knapsack, if possible
if k >= weight:
dp[i][j][k] = max(dp[i][j][k], weight +
dp[i - 1][j][k - weight])
return dp[n][capacity1][capacity2]
arr = [8, 2, 3]
capacity1 = 10
capacity2 = 3
res = maxWeight(arr, capacity1, capacity2)
print(res)
// C# program to find the largest subset of
// the array that can fit into two knapsack
// using tabulation
using System;
using System.Collections.Generic;
class GfG {
static int maxWeight(List<int> arr, int capacity1, int capacity2) {
int n = arr.Count;
// Initialize a 3D DP array with dimensions (n+1) x
// (w1+1) x (w2+1)
int[, , ] dp = new int[n + 1, capacity1 + 1, capacity2 + 1];
// Fill the DP array iteratively
for (int i = 1; i <= n; i++) {
int weight = arr[i - 1];
for (int j = 0; j <= capacity1; j++) {
for (int k = 0; k <= capacity2; k++) {
// Option 1: Don't take the current item
dp[i, j, k] = dp[i - 1, j, k];
// Option 2: Take the current item in
// the first knapsack, if possible
if (j >= weight) {
dp[i, j, k] = Math.Max(
dp[i, j, k],
weight
+ dp[i - 1, j - weight, k]);
}
// Option 3: Take the current item in
// the second knapsack, if possible
if (k >= weight) {
dp[i, j, k] = Math.Max(
dp[i, j, k],
weight
+ dp[i - 1, j, k - weight]);
}
}
}
}
return dp[n, capacity1, capacity2];
}
static void Main() {
List<int> arr = new List<int>{ 8, 2, 3 };
int capacity1 = 10, capacity2 = 3;
int res = maxWeight(arr, capacity1, capacity2);
Console.WriteLine(res);
}
}
// JavaScript program to find the largest subset of
// the array that can fit into two knapsack
// using tabulation
function maxWeight(arr, capacity1, capacity2) {
const n = arr.length;
// Initialize a 3D DP array with dimensions (n+1) x
// (capacity1+1) x (capacity2+1)
const dp = Array.from(
{length : n + 1},
() => Array.from({length : capacity1 + 1},
() => Array(capacity2 + 1).fill(0)));
// Fill the DP array iteratively
for (let i = 1; i <= n; i++) {
const weight = arr[i - 1];
for (let j = 0; j <= capacity1; j++) {
for (let k = 0; k <= capacity2; k++) {
// Option 1: Don't take the current item
dp[i][j][k] = dp[i - 1][j][k];
// Option 2: Take the current item in the
// first knapsack, if possible
if (j >= weight) {
dp[i][j][k] = Math.max(
dp[i][j][k],
weight + dp[i - 1][j - weight][k]);
}
// Option 3: Take the current item in the
// second knapsack, if possible
if (k >= weight) {
dp[i][j][k] = Math.max(
dp[i][j][k],
weight + dp[i - 1][j][k - weight]);
}
}
}
}
return dp[n][capacity1][capacity2];
}
const arr = [ 8, 2, 3 ];
const capacity1 = 10;
const capacity2= 3;
const res = maxWeight(arr, capacity1, capacity2);
console.log(res);
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