# 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 ` `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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