Fractional Knapsack Problem

Given the weights and profits of N items, in the form of {profit, weight} put these items in a knapsack of capacity W to get the maximum total profit in the knapsack. In Fractional Knapsack, we can break items for maximizing the total value of the knapsack.

Input: arr[] = {{60, 10}, {100, 20}, {120, 30}}, W = 50
Output: 240 
Explanation: By taking items of weight 10 and 20 kg and 2/3 fraction of 30 kg. 
Hence total price will be 60+100+(2/3)(120) = 240

Input:  arr[] = {{500, 30}}, W = 10
Output: 166.667

Naive Approach: To solve the problem follow the below idea:

Try all possible subsets with all different fractions.

Time Complexity: O(2^N)
Auxiliary Space: O(N)

Fractional Knapsack Problem using Greedy algorithm:

An efficient solution is to use the Greedy approach. 

The basic idea of the greedy approach is to calculate the ratio profit/weight for each item and sort the item on the basis of this ratio. Then take the item with the highest ratio and add them as much as we can (can be the whole element or a fraction of it).

This will always give the maximum profit because, in each step it adds an element such that this is the maximum possible profit for that much weight.

Illustration:

Check the below illustration for a better understanding:

Consider the example: arr[] = {{100, 20}, {60, 10}, {120, 30}}, W = 50.

Sorting: Initially sort the array based on the profit/weight ratio. The sorted array will be {{60, 10}, {100, 20}, {120, 30}}.

Iteration:

• For i = 0, weight = 10 which is less than W. So add this element in the knapsack. profit = 60 and remaining W = 50 – 10 = 40.
• For i = 1, weight = 20 which is less than W. So add this element too. profit = 60 + 100 = 160 and remaining W = 40 – 20 = 20.
• For i = 2, weight = 30 is greater than W. So add 20/30 fraction = 2/3 fraction of the element. Therefore profit = 2/3 * 120 + 160 = 80 + 160 = 240 and remaining W becomes 0.

So the final profit becomes 240 for W = 50.

Follow the given steps to solve the problem using the above approach:

• Calculate the ratio (profit/weight) for each item.
• Sort all the items in decreasing order of the ratio.
• Initialize res = 0, curr_cap = given_cap.
• Do the following for every item i in the sorted order:
  • If the weight of the current item is less than or equal to the remaining capacity then add the value of that item into the result
  • Else add the current item as much as we can and break out of the loop.
• Return res.

Below is the implementation of the above approach:

C++

// C++ program to solve fractional Knapsack Problem

#include <bits/stdc++.h>
using namespace std;

// Structure for an item which stores weight and
// corresponding value of Item
struct Item {
    int profit, weight;

    // Constructor
    Item(int profit, int weight)
    {
        this->profit = profit;
        this->weight = weight;
    }
};

// Comparison function to sort Item
// according to profit/weight ratio
static bool cmp(struct Item a, struct Item b)
{
    double r1 = (double)a.profit / (double)a.weight;
    double r2 = (double)b.profit / (double)b.weight;
    return r1 > r2;
}

// Main greedy function to solve problem
double fractionalKnapsack(int W, struct Item arr[], int N)
{
    // Sorting Item on basis of ratio
    sort(arr, arr + N, cmp);

    double finalvalue = 0.0;

    // Looping through all items
    for (int i = 0; i < N; i++) {

        // If adding Item won't overflow,
        // add it completely
        if (arr[i].weight <= W) {
            W -= arr[i].weight;
            finalvalue += arr[i].profit;
        }

        // If we can't add current Item,
        // add fractional part of it
        else {
            finalvalue
                += arr[i].profit
                   * ((double)W / (double)arr[i].weight);
            break;
        }
    }

    // Returning final value
    return finalvalue;
}

// Driver code
int main()
{
    int W = 50;
    Item arr[] = { { 60, 10 }, { 100, 20 }, { 120, 30 } };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    cout << fractionalKnapsack(W, arr, N);
    return 0;
}

Java

// Java program to solve fractional Knapsack Problem

import java.lang.*;
import java.util.Arrays;
import java.util.Comparator;

// Greedy approach
public class FractionalKnapSack {

    // Function to get maximum value
    private static double getMaxValue(ItemValue[] arr,
                                      int capacity)
    {
        // Sorting items by profit/weight ratio;
        Arrays.sort(arr, new Comparator<ItemValue>() {
            @Override
            public int compare(ItemValue item1,
                               ItemValue item2)
            {
                double cpr1
                    = new Double((double)item1.profit
                                 / (double)item1.weight);
                double cpr2
                    = new Double((double)item2.profit
                                 / (double)item2.weight);

                if (cpr1 < cpr2)
                    return 1;
                else
                    return -1;
            }
        });

        double totalValue = 0d;

        for (ItemValue i : arr) {

            int curWt = (int)i.weight;
            int curVal = (int)i.profit;

            if (capacity - curWt >= 0) {

                // This weight can be picked whole
                capacity = capacity - curWt;
                totalValue += curVal;
            }
            else {

                // Item cant be picked whole
                double fraction
                    = ((double)capacity / (double)curWt);
                totalValue += (curVal * fraction);
                capacity
                    = (int)(capacity - (curWt * fraction));
                break;
            }
        }

        return totalValue;
    }

    // Item value class
    static class ItemValue {

        int profit, weight;

        // Item value function
        public ItemValue(int val, int wt)
        {
            this.weight = wt;
            this.profit = val;
        }
    }

    // Driver code
    public static void main(String[] args)
    {

        ItemValue[] arr = { new ItemValue(60, 10),
                            new ItemValue(100, 20),
                            new ItemValue(120, 30) };

        int capacity = 50;

        double maxValue = getMaxValue(arr, capacity);

        // Function call
        System.out.println(maxValue);
    }
}

Python3

# Structure for an item which stores weight and
# corresponding value of Item
class Item:
    def __init__(self, profit, weight):
        self.profit = profit
        self.weight = weight

# Main greedy function to solve problem
def fractionalKnapsack(W, arr):

    # Sorting Item on basis of ratio
    arr.sort(key=lambda x: (x.profit/x.weight), reverse=True)

    # Result(value in Knapsack)
    finalvalue = 0.0

    # Looping through all Items
    for item in arr:

        # If adding Item won't overflow,
        # add it completely
        if item.weight <= W:
            W -= item.weight
            finalvalue += item.profit

        # If we can't add current Item,
        # add fractional part of it
        else:
            finalvalue += item.profit * W / item.weight
            break

    # Returning final value
    return finalvalue


# Driver Code
if __name__ == "__main__":
    W = 50
    arr = [Item(60, 10), Item(100, 20), Item(120, 30)]

    # Function call
    max_val = fractionalKnapsack(W, arr)
    print(max_val)

C#

// C# program to solve fractional Knapsack Problem

using System;
using System.Collections;

class GFG {

    // Class for an item which stores weight and
    // corresponding value of Item
    class item {
        public int profit;
        public int weight;

        public item(int profit, int weight)
        {
            this.profit = profit;
            this.weight = weight;
        }
    }

    // Comparison function to sort Item according
    // to val/weight ratio
    class cprCompare : IComparer {
        public int Compare(Object x, Object y)
        {
            item item1 = (item)x;
            item item2 = (item)y;
            double cpr1 = (double)item1.profit
                          / (double)item1.weight;
            double cpr2 = (double)item2.profit
                          / (double)item2.weight;

            if (cpr1 < cpr2)
                return 1;

            return cpr1 > cpr2 ? -1 : 0;
        }
    }

    // Main greedy function to solve problem
    static double FracKnapSack(item[] items, int w)
    {

        // Sort items based on cost per units
        cprCompare cmp = new cprCompare();
        Array.Sort(items, cmp);

        // Traverse items, if it can fit,
        // take it all, else take fraction
        double totalVal = 0f;
        int currW = 0;

        foreach(item i in items)
        {
            float remaining = w - currW;

            // If the whole item can be
            // taken, take it
            if (i.weight <= remaining) {
                totalVal += (double)i.profit;
                currW += i.weight;
            }

            // dd fraction until we run out of space
            else {
                if (remaining == 0)
                    break;

                double fraction
                    = remaining / (double)i.weight;
                totalVal += fraction * (double)i.profit;
                currW += (int)(fraction * (double)i.weight);
            }
        }
        return totalVal;
    }

    // Driver code
    static void Main(string[] args)
    {
        int W = 50;
        item[] arr = { new item(60, 10), new item(100, 20),
                       new item(120, 30) };

        // Function call
        Console.WriteLine(FracKnapSack(arr, W));
    }
}

// This code is contributed by Mohamed Adel

Javascript

// JavaScript program to solve fractional Knapsack Problem

// Structure for an item which stores weight and
// corresponding value of Item
class Item {
    constructor(profit, weight) {
        this.profit = profit;
        this.weight = weight;
    }
}

// Comparison function to sort Item
// according to val/weight ratio
function cmp(a, b) {
    let r1 = a.profit / a.weight;
    let r2 = b.profit / b.weight;
    return r1 > r2;
}

// Main greedy function to solve problem
function fractionalKnapsack(W, arr) {
    // Sorting Item on basis of ratio
    arr.sort(cmp);

    let finalvalue = 0.0;

    // Looping through all items
    for (let i = 0; i < arr.length; i++) {

        // If adding Item won't overflow,
        // add it completely
        if (arr[i].weight <= W) {
            W -= arr[i].weight;
            finalvalue += arr[i].profit;
        }

        // If we can't add current Item,
        // add fractional part of it
        else {
            finalvalue += arr[i].profit * (W / arr[i].weight);
            break;
        }
    }

    // Returning final value
    return finalvalue;
}

// Driver code
let W = 50;
let arr = [new Item(60, 10), new Item(100, 20), new Item(120, 30)];

console.log(fractionalKnapsack(W, arr));

// This code is contributed by lokeshpotta20

Output

240

Time Complexity: O(N * logN)
Auxiliary Space: O(N)

This article is contributed by Utkarsh Trivedi.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.