Fractional Knapsack Problem
Given the weights and profits of N items, in the form of {profit, weight} put these items in a knapsack of capacity W to get the maximum total profit in the knapsack. In Fractional Knapsack, we can break items for maximizing the total value of the knapsack.
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) = 240Input: arr[] = {{500, 30}}, W = 10
Output: 166.667
Naive Approach: To solve the problem follow the below idea:
Try all possible subsets with all different fractions.
Time Complexity: O(2N)
Auxiliary Space: O(N)
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 profit/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 as much as we can (can be the whole element or a fraction of it).
This will always give the maximum profit because, in each step it adds an element such that this is the maximum possible profit for that much weight.
Illustration:
Check the below illustration for a better understanding:
Consider the example: arr[] = {{100, 20}, {60, 10}, {120, 30}}, W = 50.
Sorting: Initially sort the array based on the profit/weight ratio. The sorted array will be {{60, 10}, {100, 20}, {120, 30}}.
Iteration:
- For i = 0, weight = 10 which is less than W. So add this element in the knapsack. profit = 60 and remaining W = 50 – 10 = 40.
- For i = 1, weight = 20 which is less than W. So add this element too. profit = 60 + 100 = 160 and remaining W = 40 – 20 = 20.
- For i = 2, weight = 30 is greater than W. So add 20/30 fraction = 2/3 fraction of the element. Therefore profit = 2/3 * 120 + 160 = 80 + 160 = 240 and remaining W becomes 0.
So the final profit becomes 240 for W = 50.
Follow the given steps to solve the problem using the above approach:
- Calculate the ratio (profit/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 <bits/stdc++.h> using namespace std; // Structure for an item which stores weight and // corresponding value of Item struct Item { int profit, weight; // Constructor Item( int profit, int weight) { this ->profit = profit; this ->weight = weight; } }; // Comparison function to sort Item // according to profit/weight ratio static bool cmp( struct Item a, struct Item b) { double r1 = ( double )a.profit / ( double )a.weight; double r2 = ( double )b.profit / ( 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].profit; } // If we can't add current Item, // add fractional part of it else { finalvalue += arr[i].profit * (( 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.lang.*; import java.util.Arrays; import java.util.Comparator; // Greedy approach public class FractionalKnapSack { // Function to get maximum value private static double getMaxValue(ItemValue[] arr, int capacity) { // Sorting items by profit/weight ratio; Arrays.sort(arr, new Comparator<ItemValue>() { @Override public int compare(ItemValue item1, ItemValue item2) { double cpr1 = new Double(( double )item1.profit / ( double )item1.weight); double cpr2 = new Double(( double )item2.profit / ( 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.profit; if (capacity - curWt >= 0 ) { // This weight can be picked whole 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 profit, weight; // Item value function public ItemValue( int val, int wt) { this .weight = wt; this .profit = 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 , profit, weight): self .profit = profit 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.profit / 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.profit # If we can't add current Item, # add fractional part of it else : finalvalue + = item.profit * 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 profit; public int weight; public item( int profit, int weight) { this .profit = profit; 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.profit / ( double )item1.weight; double cpr2 = ( double )item2.profit / ( 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.profit; 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.profit; currW += ( int )(fraction * ( double )i.weight); } } return totalVal; } // Driver code static void Main( string [] args) { int W = 50; item[] arr = { new item(60, 10), new item(100, 20), new item(120, 30) }; // Function call Console.WriteLine(FracKnapSack(arr, W)); } } // 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(profit, weight) { this .profit = profit; this .weight = weight; } } // Comparison function to sort Item // according to val/weight ratio function cmp(a, b) { let r1 = a.profit / a.weight; let r2 = b.profit / 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].profit; } // If we can't add current Item, // add fractional part of it else { finalvalue += arr[i].profit * (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 |
240
Time Complexity: O(N * logN)
Auxiliary Space: O(N)
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