Maximum sum of values of N items in 0-1 Knapsack by reducing weight of at most K items in half
Given weights and values of N items and the capacity W of the knapsack. Also given that the weight of at most K items can be changed to half of its original weight. The task is to find the maximum sum of values of N items that can be obtained such that the sum of weights of items in knapsack does not exceed the given capacity W.
Examples:
Input: W = 4, K = 1, value = [17, 20, 10, 15], weight = [4, 2, 7, 5]
Output: 37
Explanation: Change the weight of at most K items to half of the weight in a optimal way to get maximum value. Decrease the weight of first item to half and add second item weight the resultant sum of value is 37 which is maximumInput: W = 8, K = 2, value = [17, 20, 10, 15], weight = [4, 2, 7, 5]
Output: 53
Explanation: Change the weight of the last item and first item and the add the weight the of the 2nd item, The total sum value of item will be 53.
Approach: Given problem is the variation of the 0 1 knapsack problem. Flag indicates number of items whose weight has been reduced to half. At every recursive call maximum of following cases is calculated and returned:
- Base case: If the index exceeds the length of values then return zero
- If flag is equal to K, maximum of 2 cases is considered:
- Include item with full weight if item’s weight does not exceed remaining weight
- Skip the item
- If flag is less than K, maximum of 3 cases is considered:
- Include item with full weight if item’s weight does not exceed remaining weight
- Include item with half weight if item’s half weight does not exceed remaining weight
- Skip the item
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find the maximum valueint maximum(int value[], int weight[], int weight1, int flag, int K, int index, int val_len){ // base condition if (index >= val_len) { return 0; } // K elements already reduced // to half of their weight if (flag == K) { // Dont include item int skip = maximum(value, weight, weight1, flag, K, index + 1, val_len); int full = 0; // If weight of the item is // less than or equal to the // remaining weight then include // the item if (weight[index] <= weight1) { full = value[index] + maximum( value, weight, weight1 - weight[index], flag, K, index + 1, val_len); } // Return the maximum of // both cases return max(full, skip); } // If the weight reduction to half // is possible else { // Skip the item int skip = maximum( value, weight, weight1, flag, K, index + 1, val_len); int full = 0; int half = 0; // Include item with full weight // if weight of the item is less // than the remaining weight if (weight[index] <= weight1) { full = value[index] + maximum( value, weight, weight1 - weight[index], flag, K, index + 1, val_len); } // Include item with half weight // if half weight of the item is // less than the remaining weight if (weight[index] / 2 <= weight1) { half = value[index] + maximum( value, weight, weight1 - weight[index] / 2, flag, K, index + 1, val_len); } // Return the maximum of all 3 cases return max(full, max(skip, half)); }}int main(){ int value[] = {17, 20, 10, 15}; int weight[] = {4, 2, 7, 5}; int K = 1; int W = 4; int val_len = sizeof(value) / sizeof(value[0]); cout << (maximum(value, weight, W, 0, K, 0, val_len)); return 0;}// This code is contributed by Potta Lokesh |
Java
// Java implementation for the above approachimport java.io.*;import java.util.*;class GFG { // Function to find the maximum value static int maximum(int value[], int weight[], int weight1, int flag, int K, int index) { // base condition if (index >= value.length) { return 0; } // K elements already reduced // to half of their weight if (flag == K) { // Dont include item int skip = maximum(value, weight, weight1, flag, K, index + 1); int full = 0; // If weight of the item is // less than or equal to the // remaining weight then include // the item if (weight[index] <= weight1) { full = value[index] + maximum( value, weight, weight1 - weight[index], flag, K, index + 1); } // Return the maximum of // both cases return Math.max(full, skip); } // If the weight reduction to half // is possible else { // Skip the item int skip = maximum( value, weight, weight1, flag, K, index + 1); int full = 0; int half = 0; // Include item with full weight // if weight of the item is less // than the remaining weight if (weight[index] <= weight1) { full = value[index] + maximum( value, weight, weight1 - weight[index], flag, K, index + 1); } // Include item with half weight // if half weight of the item is // less than the remaining weight if (weight[index] / 2 <= weight1) { half = value[index] + maximum( value, weight, weight1 - weight[index] / 2, flag, K, index + 1); } // Return the maximum of all 3 cases return Math.max(full, Math.max(skip, half)); } } public static void main(String[] args) throws Exception { int value[] = { 17, 20, 10, 15 }; int weight[] = { 4, 2, 7, 5 }; int K = 1; int W = 4; System.out.println( maximum(value, weight, W, 0, K, 0)); }} |
Python3
# Python program for the above approach# Function to find the maximum valuedef maximum(value, weight, weight1, flag, K, index, val_len) : # base condition if (index >= val_len) : return 0 # K elements already reduced # to half of their weight if (flag == K) : # Dont include item skip = maximum(value, weight, weight1, flag, K, index + 1, val_len) full = 0 # If weight of the item is # less than or equal to the # remaining weight then include # the item if (weight[index] <= weight1) : full = value[index] + maximum( value, weight, weight1 - weight[index], flag, K, index + 1, val_len) # Return the maximum of # both cases return max(full, skip) # If the weight reduction to half # is possible else : # Skip the item skip = maximum( value, weight, weight1, flag, K, index + 1, val_len) full = 0 half = 0 # Include item with full weight # if weight of the item is less # than the remaining weight if (weight[index] <= weight1) : full = value[index] + maximum( value, weight, weight1 - weight[index], flag, K, index + 1, val_len) # Include item with half weight # if half weight of the item is # less than the remaining weight if (weight[index] / 2 <= weight1) : half = value[index] + maximum( value, weight, weight1 - weight[index] / 2, flag, K, index + 1, val_len) # Return the maximum of all 3 cases return max(full, max(skip, half)) # Driver Codevalue = [ 17, 20, 10, 15 ]weight = [ 4, 2, 7, 5 ]K = 1W = 4val_len = len(value)print(maximum(value, weight, W, 0, K, 0, val_len))# This code is contributed by sanjoy_62. |
C#
// C# implementation for the above approachusing System;public class GFG { // Function to find the maximum value static int maximum(int []value, int []weight, int weight1, int flag, int K, int index) { // base condition if (index >= value.Length) { return 0; } // K elements already reduced // to half of their weight if (flag == K) { // Dont include item int skip = maximum(value, weight, weight1, flag, K, index + 1); int full = 0; // If weight of the item is // less than or equal to the // remaining weight then include // the item if (weight[index] <= weight1) { full = value[index] + maximum( value, weight, weight1 - weight[index], flag, K, index + 1); } // Return the maximum of // both cases return Math.Max(full, skip); } // If the weight reduction to half // is possible else { // Skip the item int skip = maximum( value, weight, weight1, flag, K, index + 1); int full = 0; int half = 0; // Include item with full weight // if weight of the item is less // than the remaining weight if (weight[index] <= weight1) { full = value[index] + maximum( value, weight, weight1 - weight[index], flag, K, index + 1); } // Include item with half weight // if half weight of the item is // less than the remaining weight if (weight[index] / 2 <= weight1) { half = value[index] + maximum( value, weight, weight1 - weight[index] / 2, flag, K, index + 1); } // Return the maximum of all 3 cases return Math.Max(full, Math.Max(skip, half)); } } // Driver code public static void Main(String[] args) { int []value = { 17, 20, 10, 15 }; int []weight = { 4, 2, 7, 5 }; int K = 1; int W = 4; Console.WriteLine( maximum(value, weight, W, 0, K, 0)); }}// This code is contributed by shikhasingrajput |
Javascript
<script>// javascript implementation for the above approach // Function to find the maximum value function maximum(value, weight , weight1, flag , K , index) { // base condition if (index >= value.length) { return 0; } // K elements already reduced // to half of their weight if (flag == K) { // Dont include item var skip = maximum(value, weight, weight1, flag, K, index + 1); var full = 0; // If weight of the item is // less than or equal to the // remaining weight then include // the item if (weight[index] <= weight1) { full = value[index] + maximum( value, weight, weight1 - weight[index], flag, K, index + 1); } // Return the maximum of // both cases return Math.max(full, skip); } // If the weight reduction to half // is possible else { // Skip the item var skip = maximum( value, weight, weight1, flag, K, index + 1); var full = 0; var half = 0; // Include item with full weight // if weight of the item is less // than the remaining weight if (weight[index] <= weight1) { full = value[index] + maximum( value, weight, weight1 - weight[index], flag, K, index + 1); } // Include item with half weight // if half weight of the item is // less than the remaining weight if (weight[index] / 2 <= weight1) { half = value[index] + maximum( value, weight, weight1 - weight[index] / 2, flag, K, index + 1); } // Return the maximum of all 3 cases return Math.max(full, Math.max(skip, half)); } }// Driver codevar value = [ 17, 20, 10, 15 ];var weight = [ 4, 2, 7, 5 ];var K = 1;var W = 4;document.write( maximum(value, weight, W, 0, K, 0));// This code is contributed by Princi Singh </script> |
Output
37
Time Complexity: O(3^N)
Auxiliary Space: O(N)
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