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: 244.444Input: 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 valuefloat 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 codeint 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 approachclass 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 approachimport sys # Function to return the maximum required valuedef 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 codeval = [ 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 approachusing 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 valuefunction 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 codevar 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.
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