Unbounded Fractional Knapsack

Given 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 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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea here is to just find the item which has 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 implementataion 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 

Output:

244.444

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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