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# Fractional Knapsack Problem

• Difficulty Level : Medium
• Last Updated : 04 Feb, 2023

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

Note: In the 0-1 Knapsack problem, we are not allowed to break items. We either take the whole item or don’t take it.

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

Recommended Practice

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

Try all possible subsets with all different fractions.

## Algorithm for Knapsack problem using Greedy Strategy:

Algorithm Fr-Knapsack (M,n)
/*Input: Knapsack capacity = M. n is the number of available items with associated profits P[1: n] and weights W[1: n]. These items are such that P[i]/W[i]>=P[i+1]/W[i +1] where 1 â‰¤iâ‰¤n.

Output: S[1: n] is the fixed size solution vector. S[i] gives the fractional part xi, of item i placed into a knapsack, 0 â‰¤ xi â‰¤1 and 1 â‰¤ i â‰¤ n. */
{
for (i:=1; i <= n ;i++)

{

S[i] :=0.0; // Initialization of a solution vector.
}

balance:= M; /* 'balance' describes the remaining capacity of a knapsack. Initially it is equal to the knapsack capacity M. */

for(i:=1; i <= n ;i++)

{
if(W [i] > balance ) / *Insufficient capacity of a knapsack */
break;

S[i] :=1.0; // add whole item i.
balance := balance - W[i]; /*Update the remaining capacity after adding item i with weight W[i] into a knapsack.*/
}
f(i <= n) /*Due to insufficient capacity add fractional part of item i in the knapsack. */
S[i]:= (balance / W[i] ) ;
return S;
}

## 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 value/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 until we canâ€™t add the next item as a whole and at the end add the next item as much as we can. Which will always be the optimal solution to this problem.

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

• Calculate the ratio(value/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 using namespace std;   // Structure for an item which stores weight and // corresponding value of Item struct Item {     int value, weight;       // Constructor     Item(int value, int weight)     {         this->value = value;         this->weight = weight;     } };   // Comparison function to sort Item // according to val/weight ratio static bool cmp(struct Item a, struct Item b) {     double r1 = (double)a.value / (double)a.weight;     double r2 = (double)b.value / (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].value;         }           // If we can't add current Item,         // add fractional part of it         else {             finalvalue                 += arr[i].value                    * ((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.io.*; import java.util.Arrays; import java.util.Comparator;   // Greedy approach class FractionalKnapSack {           // Function to get maximum value     private static double getMaxValue(ItemValue[] arr,                                       int capacity)     {         // Sorting items by value/weight ratio;         Arrays.sort(arr, new Comparator() {             @Override             public int compare(ItemValue item1,                                ItemValue item2)             {                 double cpr1                     = new Double((double)item1.value                                  / (double)item1.weight);                 double cpr2                     = new Double((double)item2.value                                  / (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.value;               if (capacity - curWt >= 0) {                   // this weight can be picked while                 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 value, weight;           // Item value function         public ItemValue(int val, int wt)         {             this.weight = wt;             this.value = 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, value, weight):         self.value = value         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.value/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.value           # If we can't add current Item,         # add fractional part of it         else:             finalvalue += item.value * 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 value;         public int weight;           public item(int value, int weight)         {             this.value = value;             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.value                           / (double)item1.weight;             double cpr2 = (double)item2.value                           / (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.value;                 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.value;                 currW += (int)(fraction * (double)i.weight);             }         }         return totalVal;     }       // Driver code     static void Main(string[] args)     {         item[] arr = { new item(60, 10), new item(100, 20),                        new item(120, 30) };           // Function call         Console.WriteLine(FracKnapSack(arr, 50));     } }   // 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(value, weight) {         this.value = value;         this.weight = weight;     } }   // Comparison function to sort Item // according to val/weight ratio function cmp(a, b) {     let r1 = a.value / a.weight;     let r2 = b.value / 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].value;         }           // If we can't add current Item,         // add fractional part of it         else {             finalvalue += arr[i].value * (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

Maximum value we can obtain = 240

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