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Unbounded Fractional Knapsack

  • Difficulty Level : Easy
  • Last Updated : 27 May, 2021

Given the weights and values of n items, the task is to put these items in a knapsack of capacity W to get the maximum total value in the knapsack, we can repeatedly put the same item and we can also put a fraction of an item.

Examples:  

Input: val[] = {14, 27, 44, 19}, wt[] = {6, 7, 9, 8}, W = 50 
Output: 244.444

Input: val[] = {100, 60, 120}, wt[] = {20, 10, 30}, W = 50 
Output: 300 

Approach: The idea here is to just find the item which has the largest value to weight ratio. Then fill the whole knapsack with this item only, in order to maximize the final value of the knapsack.



Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the maximum required value
float knapSack(int W, float wt[], float val[], int n)
{
 
    // maxratio will store the maximum value to weight
    // ratio we can have for any item and maxindex
    // will store the index of that element
    float maxratio = INT_MIN;
    int maxindex = 0;
 
    // Find the maximum ratio
    for (int i = 0; i < n; i++) {
        if ((val[i] / wt[i]) > maxratio) {
            maxratio = (val[i] / wt[i]);
            maxindex = i;
        }
    }
 
    // The item with the maximum value to
    // weight ratio will be put into
    // the knapsack repeatedly until full
    return (W * maxratio);
}
 
// Driver code
int main()
{
    float val[] = { 14, 27, 44, 19 };
    float wt[] = { 6, 7, 9, 8 };
    int n = sizeof(val) / sizeof(val[0]);
    int W = 50;
 
    cout << knapSack(W, wt, val, n);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
 
    // Function to return the maximum required value
    static float knapSack(int W, float wt[],
                            float val[], int n)
    {
 
        // maxratio will store the maximum value to weight
        // ratio we can have for any item and maxindex
        // will store the index of that element
        float maxratio = Integer.MIN_VALUE;
        int maxindex = 0;
 
        // Find the maximum ratio
        for (int i = 0; i < n; i++)
        {
            if ((val[i] / wt[i]) > maxratio)
            {
                maxratio = (val[i] / wt[i]);
                maxindex = i;
            }
        }
 
        // The item with the maximum value to
        // weight ratio will be put into
        // the knapsack repeatedly until full
        return (W * maxratio);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        float val[] = {14, 27, 44, 19};
        float wt[] = {6, 7, 9, 8};
        int n = val.length;
        int W = 50;
 
        System.out.println(knapSack(W, wt, val, n));
    }
}
 
// This code is contributed by 29AjayKumar

Python3




# Python implementation of the approach
 
import sys
 
# Function to return the maximum required value
def knapSack(W, wt, val, n):
 
    # maxratio will store the maximum value to weight
    # ratio we can have for any item and maxindex
    # will store the index of that element
    maxratio = -sys.maxsize-1;
    maxindex = 0;
 
    # Find the maximum ratio
    for i in range(n):
        if ((val[i] / wt[i]) > maxratio):
            maxratio = (val[i] / wt[i]);
            maxindex = i;
 
 
    # The item with the maximum value to
    # weight ratio will be put into
    # the knapsack repeatedly until full
    return (W * maxratio);
 
 
# Driver code
 
val = [ 14, 27, 44, 19 ];
wt = [ 6, 7, 9, 8 ];
n = len(val);
W = 50;
 
print(knapSack(W, wt, val, n));
 
# This code is contributed by Rajput-Ji

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the maximum required value
    static float knapSack(int W, float []wt,
                            float []val, int n)
    {
 
        // maxratio will store the maximum value to weight
        // ratio we can have for any item and maxindex
        // will store the index of that element
        float maxratio = int.MinValue;
        int maxindex = 0;
 
        // Find the maximum ratio
        for (int i = 0; i < n; i++)
        {
            if ((val[i] / wt[i]) > maxratio)
            {
                maxratio = (val[i] / wt[i]);
                maxindex = i;
            }
        }
 
        // The item with the maximum value to
        // weight ratio will be put into
        // the knapsack repeatedly until full
        return (W * maxratio);
    }
 
    // Driver code
    public static void Main()
    {
        float []val = {14, 27, 44, 19};
        float []wt = {6, 7, 9, 8};
        int n = val.Length;
        int W = 50;
 
        Console.WriteLine(knapSack(W, wt, val, n));
    }
}
 
// This code is contributed by AnkitRai01

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the maximum required value
function knapSack(W, wt, val, n)
{
 
    // maxratio will store the maximum value to weight
    // ratio we can have for any item and maxindex
    // will store the index of that element
    var maxratio = -1000000000;
    var maxindex = 0;
 
    // Find the maximum ratio
    for (var i = 0; i < n; i++) {
        if (parseInt(val[i] / wt[i]) > maxratio) {
            maxratio = (val[i] / wt[i]);
            maxindex = i;
        }
    }
 
    // The item with the maximum value to
    // weight ratio will be put into
    // the knapsack repeatedly until full
    return (W * maxratio);
}
 
// Driver code
var val = [14, 27, 44, 19];
var wt = [6, 7, 9, 8];
var n = val.length;
var W = 50;
document.write( knapSack(W, wt, val, n).toFixed(3));
 
</script>
Output: 
244.444

 

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