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++ 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 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 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# 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 |
<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> |
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++ implementation of the tabulation 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];
// Function to calculate // maximum weight that can be put // into the knapsacks 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] =
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 code int 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;
} |
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));
}
} |
# Constants for maximum values maxN = 31
maxW = 31
# 3D array to store states of DP dp = [[[ 0 for _ in range (maxW)] for _ in range (maxW)] for _ in range (maxN)]
# Function to calculate maximum weight that can be put into the knapsacks def 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 code if __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))
|
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));
}
} |
const maxN = 31; const maxW = 31; // 3D array to store states of DP const 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 knapsacks function 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 array const arr = [8, 2, 3]; // Number of elements in the array const n = arr.length; // Capacity of knapsacks const w1 = 10, w2 = 3; // Function to be called console.log(maxWeight(arr, n, w1, w2)); |
13
Time complexity: O(N*W1*W2)
Auxiliary Space: O(N*W1*W2))