Unbounded Fractional Knapsack

Last Updated : 14 Apr, 2023

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# 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

Time Complexity: O(n) where n is size of input array val and wt. This is because a for loop is being executed from 1 till n in knapSack function.

Space Complexity: O(1) as no extra space has been used.

Unbounded Knapsack (Repetition of items allowed)

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Unbounded Fractional Knapsack

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