Knapsack with large Weights

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Given a knapsack with capacity C and two arrays w[] and val[] representing the weights and values of N distinct items, the task is to find the maximum value you can put into the knapsack. Items cannot be broken and an item with weight X takes X capacity of the knapsack.

Examples:

Input: w[] = {3, 4, 5}, val[] = {30, 50, 60}, C = 8
Output: 90
We take objects ‘1’ and ‘3’.
The total value we get is (30 + 60) = 90.
Total weight is 8, thus it fits in the given capacity

Input: w[] = {10000}, val[] = {10}, C = 100000
Output: 10

Approach: The traditional famous 0-1 knapsack problem can be solved in O(N*C) time but if the capacity of the knapsack is huge then a 2D N*C array can’t make be made. Luckily, it can be solved by redefining the states of the dp.
Let’s have a look at the states of the DP first.
dp[V][i] represents the minimum weight subset of the subarray arr[i…N-1] required to get a value of at least V. The recurrence relation will be:

dp[V][i] = min(dp[V][i+1], w[i] + dp[V – val[i]][i + 1])

So, for each V from 0 to the maximum value of V possible, try to find if the given V can be represented with the given array. The largest such V that can be represented becomes the required answer.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
#define V_SUM_MAX 1000
#define N_MAX 100
#define W_MAX 10000000
 
// To store the states of DP
int dp[V_SUM_MAX + 1][N_MAX];
bool v[V_SUM_MAX + 1][N_MAX];
 
// Function to solve the recurrence relation
int solveDp(int r, int i, vector<int>& w, vector<int>& val, int n)
{
    // Base cases
    if (r <= 0)
        return 0;
    if (i == n)
        return W_MAX;
    if (v[r][i])
        return dp[r][i];
 
    // Marking state as solved
    v[r][i] = 1;
 
    // Recurrence relation
    dp[r][i]
        = min(solveDp(r, i + 1, w, val, n),
              w[i] + solveDp(r - val[i],
                             i + 1, w, val, n));
    return dp[r][i];
}
 
// Function to return the maximum weight
int maxWeight(vector<int>& w, vector<int>& val, int n, int c)
{
 
    // Iterating through all possible values
    // to find the largest value that can
    // be represented by the given weights
    for (int i = V_SUM_MAX; i >= 0; i--) {
        if (solveDp(i, 0, w, val, n) <= c) {
            return i;
        }
    }
    return 0;
}
 
// Driver code
int main()
{
    vector<int> w = { 3, 4, 5 };
    vector<int> val = { 30, 50, 60 };
    int n = (int)w.size();
    int C = 8;
 
    cout << maxWeight(w, val, n, C);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG 
{
    static final int V_SUM_MAX = 1000;
    static final int N_MAX = 100;
    static final int W_MAX = 10000000;
     
    // To store the states of DP 
    static int dp[][] = new int[V_SUM_MAX + 1][N_MAX]; 
    static boolean v[][] = new boolean [V_SUM_MAX + 1][N_MAX]; 
     
    // Function to solve the recurrence relation 
    static int solveDp(int r, int i, int w[],       
                          int val[], int n) 
    { 
        // Base cases 
        if (r <= 0) 
            return 0; 
             
        if (i == n) 
            return W_MAX; 
             
        if (v[r][i]) 
            return dp[r][i]; 
     
        // Marking state as solved 
        v[r][i] = true; 
     
        // Recurrence relation 
        dp[r][i] = Math.min(solveDp(r, i + 1, w, val, n), 
                     w[i] + solveDp(r - val[i], 
                                    i + 1, w, val, n)); 
         
        return dp[r][i]; 
    } 
     
    // Function to return the maximum weight 
    static int maxWeight(int w[], int val[], 
                         int n, int c) 
    { 
     
        // Iterating through all possible values 
        // to find the largest value that can 
        // be represented by the given weights 
        for (int i = V_SUM_MAX; i >= 0; i--)
        { 
            if (solveDp(i, 0, w, val, n) <= c) 
            { 
                return i; 
            } 
        } 
        return 0; 
    } 
     
    // Driver code 
    public static void main (String[] args)
    { 
        int w[] = { 3, 4, 5 }; 
        int val[] = { 30, 50, 60 }; 
        int n = w.length; 
        int C = 8; 
     
        System.out.println(maxWeight(w, val, n, C)); 
    } 
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
V_SUM_MAX = 1000
N_MAX = 100
W_MAX = 10000000
 
# To store the states of DP
dp = [[ 0 for i in range(N_MAX)]
          for i in range(V_SUM_MAX + 1)]
v = [[ 0 for i in range(N_MAX)]
         for i in range(V_SUM_MAX + 1)]
 
# Function to solve the recurrence relation
def solveDp(r, i, w, val, n):
     
    # Base cases
    if (r <= 0):
        return 0
    if (i == n):
        return W_MAX
    if (v[r][i]):
        return dp[r][i]
 
    # Marking state as solved
    v[r][i] = 1
 
    # Recurrence relation
    dp[r][i] = min(solveDp(r, i + 1, w, val, n), 
            w[i] + solveDp(r - val[i], i + 1,
                            w, val, n))
    return dp[r][i]
 
# Function to return the maximum weight
def maxWeight( w, val, n, c):
 
    # Iterating through all possible values
    # to find the largest value that can
    # be represented by the given weights
    for i in range(V_SUM_MAX, -1, -1):
        if (solveDp(i, 0, w, val, n) <= c):
            return i
 
    return 0
 
# Driver code
if __name__ == '__main__':
    w = [3, 4, 5]
    val = [30, 50, 60]
    n = len(w)
    C = 8
 
    print(maxWeight(w, val, n, C))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
 
class GFG 
{
    static readonly int V_SUM_MAX = 1000;
    static readonly int N_MAX = 100;
    static readonly int W_MAX = 10000000;
     
    // To store the states of DP 
    static int [,]dp = new int[V_SUM_MAX + 1, N_MAX]; 
    static bool [,]v = new bool [V_SUM_MAX + 1, N_MAX]; 
     
    // Function to solve the recurrence relation 
    static int solveDp(int r, int i, int []w,     
                       int []val, int n) 
    { 
        // Base cases 
        if (r <= 0) 
            return 0; 
             
        if (i == n) 
            return W_MAX; 
             
        if (v[r, i]) 
            return dp[r, i]; 
     
        // Marking state as solved 
        v[r, i] = true; 
     
        // Recurrence relation 
        dp[r, i] = Math.Min(solveDp(r, i + 1, w, val, n), 
                     w[i] + solveDp(r - val[i], 
                                    i + 1, w, val, n)); 
         
        return dp[r, i]; 
    } 
     
    // Function to return the maximum weight 
    static int maxWeight(int []w, int []val, 
                         int n, int c) 
    { 
     
        // Iterating through all possible values 
        // to find the largest value that can 
        // be represented by the given weights 
        for (int i = V_SUM_MAX; i >= 0; i--)
        { 
            if (solveDp(i, 0, w, val, n) <= c) 
            { 
                return i; 
            } 
        } 
        return 0; 
    } 
     
    // Driver code 
    public static void Main(String[] args)
    { 
        int []w = { 3, 4, 5 }; 
        int []val = { 30, 50, 60 }; 
        int n = w.Length; 
        int C = 8; 
     
        Console.WriteLine(maxWeight(w, val, n, C)); 
    } 
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// Javascript implementation of the approach
 
var V_SUM_MAX = 1000
var N_MAX = 100
var W_MAX = 10000000
 
// To store the states of DP
var dp = Array.from(Array(V_SUM_MAX+1), ()=> Array(N_MAX));
var v = Array.from(Array(V_SUM_MAX+1), ()=> Array(N_MAX));
 
// Function to solve the recurrence relation
function solveDp(r, i, w, val, n)
{
    // Base cases
    if (r <= 0)
        return 0;
    if (i == n)
        return W_MAX;
    if (v[r][i])
        return dp[r][i];
 
    // Marking state as solved
    v[r][i] = 1;
 
    // Recurrence relation
    dp[r][i]
        = Math.min(solveDp(r, i + 1, w, val, n),
              w[i] + solveDp(r - val[i],
                             i + 1, w, val, n));
    return dp[r][i];
}
 
// Function to return the maximum weight
function maxWeight(w, val, n, c)
{
 
    // Iterating through all possible values
    // to find the largest value that can
    // be represented by the given weights
    for (var i = V_SUM_MAX; i >= 0; i--) {
        if (solveDp(i, 0, w, val, n) <= c) {
            return i;
        }
    }
    return 0;
}
 
// Driver code
var w = [3, 4, 5];
var val = [30, 50, 60];
var n = w.length;
var C = 8;
document.write( maxWeight(w, val, n, C));
 
</script>

Output:

Time Complexity: O(V_sum * N) where V_sum is the sum of all the values in the array val[].
Auxiliary Space : O(V_sum * N) where V_sum is the sum of all the values in the array val[].

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 table DP to store the solution of the subproblems.
Initialize the table with base cases
Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
At last get the maximum value from DP and return the final solution.

Implementation :

C++

// C++ program for above approach
 
#include <bits/stdc++.h>
using namespace std;
 
#define V_SUM_MAX 1000
#define N_MAX 100
#define W_MAX 10000000
 
// Function to return the maximum weight
int maxWeight(vector<int>& w, vector<int>& val, int n,
              int c)
{
    // Initialize dp array
    int dp[V_SUM_MAX + 1][N_MAX + 1];
    for (int i = 0; i <= V_SUM_MAX; i++)
        for (int j = 0; j <= n; j++)
            dp[i][j] = W_MAX;
 
    // Base case initialization
    for (int i = 0; i <= n; i++)
        dp[0][i] = 0;
 
    // iterate over subproblems ans get
    // the current value from previous computation
    for (int i = 1; i <= V_SUM_MAX; i++)
        for (int j = 1; j <= n; j++)
            dp[i][j] = min(
                dp[i][j - 1],
                (i >= val[j - 1])
                    ? w[j - 1] + dp[i - val[j - 1]][j - 1]
                    : W_MAX);
 
    // Finding maximum value
    for (int i = V_SUM_MAX; i >= 0; i--)
        if (dp[i][n] <= c)
            return i;
 
    return 0;
}
 
// Driver code
int main()
{
    vector<int> w = { 3, 4, 5 };
    vector<int> val = { 30, 50, 60 };
    int n = (int)w.size();
    int C = 8;
 
    cout << maxWeight(w, val, n, C);
 
    return 0;
}

Python3

import sys
 
V_SUM_MAX = 1000
N_MAX = 100
W_MAX = 10000000
 
# Function to return the maximum weight
 
 
def maxWeight(w, val, n, c):
    # Initialize dp array
    dp = [[W_MAX for j in range(n+1)] for i in range(V_SUM_MAX+1)]
 
    # Base case initialization
    for i in range(n+1):
        dp[0][i] = 0
 
    # iterate over subproblems ans get
    # the current value from previous computation
    for i in range(1, V_SUM_MAX+1):
        for j in range(1, n+1):
            dp[i][j] = min(dp[i][j-1], w[j-1]+dp[i-val[j-1]]
                           [j-1] if i >= val[j-1] else W_MAX)
 
    # Finding maximum value
    for i in range(V_SUM_MAX, -1, -1):
        if dp[i][n] <= c:
            return i
 
    return 0
 
 
# Driver code
if __name__ == "__main__":
    w = [3, 4, 5]
    val = [30, 50, 60]
    n = len(w)
    C = 8
 
    print(maxWeight(w, val, n, C))

Java

import java.util.*;
 
class Main {
    public static final int V_SUM_MAX = 1000;
    public static final int N_MAX = 100;
    public static final int W_MAX = 10000000;
 
    // Function to return the maximum weight
    public static int maxWeight(List<Integer> w,
                                List<Integer> val, int n,
                                int c)
    {
        // Initialize dp array
        int[][] dp = new int[V_SUM_MAX + 1][N_MAX + 1];
        for (int i = 0; i <= V_SUM_MAX; i++) {
            for (int j = 0; j <= n; j++) {
                dp[i][j] = W_MAX;
            }
        }
 
        // Base case initialization
        for (int i = 0; i <= n; i++) {
            dp[0][i] = 0;
        }
 
        // iterate over subproblems ans get
        // the current value from previous computation
        for (int i = 1; i <= V_SUM_MAX; i++) {
            for (int j = 1; j <= n; j++) {
                dp[i][j] = Math.min(
                    dp[i][j - 1],
                    (i >= val.get(j - 1))
                        ? w.get(j - 1)
                              + dp[i - val.get(j - 1)]
                                  [j - 1]
                        : W_MAX);
            }
        }
 
        // Finding maximum value
        for (int i = V_SUM_MAX; i >= 0; i--) {
            if (dp[i][n] <= c) {
                return i;
            }
        }
 
        return 0;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        List<Integer> w
            = new ArrayList<>(Arrays.asList(3, 4, 5));
        List<Integer> val
            = new ArrayList<>(Arrays.asList(30, 50, 60));
        int n = w.size();
        int C = 8;
 
        System.out.println(maxWeight(w, val, n, C));
    }
}

C#

using System;
 
public class KnapsackMaxWeight
{
    private const int V_SUM_MAX = 1000;
    private const int N_MAX = 100;
    private const int W_MAX = 10000000;
 
    public static int MaxWeight(int[] w, int[] val, int n, int c)
    {
        // Initialize dp array
        int[,] dp = new int[V_SUM_MAX + 1, N_MAX + 1];
        for (int i = 0; i <= V_SUM_MAX; i++)
        {
            for (int j = 0; j <= n; j++)
            {
                dp[i, j] = W_MAX;
            }
        }
 
        // Base case initialization
        for (int i = 0; i <= n; i++)
        {
            dp[0, i] = 0;
        }
 
        // iterate over subproblems ans get
        // the current value from previous computation
        for (int i = 1; i <= V_SUM_MAX; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                dp[i, j] = Math.Min(
                    dp[i, j - 1],
                    (i >= val[j - 1])
                        ? w[j - 1] + dp[i - val[j - 1], j - 1]
                        : W_MAX);
            }
        }
 
        // Finding maximum value
        for (int i = V_SUM_MAX; i >= 0; i--)
        {
            if (dp[i, n] <= c)
            {
                return i;
            }
        }
 
        return 0;
    }
 
    public static void Main()
    {
        int[] w = { 3, 4, 5 };
        int[] val = { 30, 50, 60 };
        int n = w.Length;
        int c = 8;
 
        Console.WriteLine(MaxWeight(w, val, n, c));
    }
}

Javascript

function MaxWeight(w, val, n, c) {
  // Initialize dp array
  const V_SUM_MAX = 1000;
  const N_MAX = 100;
  const W_MAX = 10000000;
  const dp = new Array(V_SUM_MAX + 1).fill().map(() => new Array(N_MAX + 1).fill(W_MAX));
 
  // Base case initialization
  for (let i = 0; i <= n; i++) {
    dp[0][i] = 0;
  }
 
  // iterate over subproblems and get
  // the current value from previous computation
  for (let i = 1; i <= V_SUM_MAX; i++) {
    for (let j = 1; j <= n; j++) {
      dp[i][j] = Math.min(
        dp[i][j - 1],
        (i >= val[j - 1]) ? w[j - 1] + dp[i - val[j - 1]][j - 1] : W_MAX
      );
    }
  }
 
  // Finding maximum value
  for (let i = V_SUM_MAX; i >= 0; i--) {
    if (dp[i][n] <= c) {
      return i;
    }
  }
 
  return 0;
}
 
const w = [3, 4, 5];
const val = [30, 50, 60];
const n = w.length;
const c = 8;
 
console.log(MaxWeight(w, val, n, c));

Output

Time Complexity: O(V_sum * N).
Auxiliary Space : O(V_sum * N).

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