Given weights and profits of N items, put these items in a knapsack of capacity W. The task is to print all possible solutions to the problem in such a way that there are no remaining items left whose weight is less than the remaining capacity of the knapsack. Also, compute the maximum profit.
Examples:
Input: Profits[] = {60, 100, 120, 50}
Weights[] = {10, 20, 30, 40}, W = 40
Output:
10: 60, 20: 100,
10: 60, 30: 120,
Maximum Profit = 180
Explanation:
Maximum profit from all the possible solutions is 180Input: Profits[] = {60, 100, 120, 50}
Weights[] = {10, 20, 30, 40}, W = 50
Output:
10: 60, 20: 100,
10: 60, 30: 120,
20: 100, 30: 120,
Maximum Profit = 220
Explanation:
Maximum profit from all the possible solutions is 220
Approach: The idea is to make pairs for the weight and the profits of the items and then try out all permutations of the array and including the weights until their is no such item whose weight is less than the remaining capacity of the knapsack. Meanwhile after including an item increment the profit for that solution by the profit of that item.
Below is the implementation of the above approach:
C++
// C++ implementation to print all// the possible solutions of the// 0/1 Knapsack problem#include <bits/stdc++.h>using namespace std;// Utility function to find the// maximum of the two elementsint max(int a, int b) { return (a > b) ? a : b; }// Function to find the all the// possible solutions of the // 0/1 knapSack problemint knapSack(int W, vector<int> wt, vector<int> val, int n){ // Mapping weights with Profits map<int, int> umap; set<vector<pair<int, int>>> set_sol; // Making Pairs and inserting // into the map for (int i = 0; i < n; i++) { umap.insert({ wt[i], val[i] }); } int result = INT_MIN; int remaining_weight; int sum = 0; // Loop to iterate over all the // possible permutations of array do { sum = 0; // Initially bag will be empty remaining_weight = W; vector<pair<int, int>> possible; // Loop to fill up the bag // until there is no weight // such which is less than // remaining weight of the // 0-1 knapSack for (int i = 0; i < n; i++) { if (wt[i] <= remaining_weight) { remaining_weight -= wt[i]; auto itr = umap.find(wt[i]); sum += (itr->second); possible.push_back({itr->first, itr->second }); } } sort(possible.begin(), possible.end()); if (sum > result) { result = sum; } if (set_sol.find(possible) == set_sol.end()){ for (auto sol: possible){ cout << sol.first << ": " << sol.second << ", "; } cout << endl; set_sol.insert(possible); } } while ( next_permutation(wt.begin(), wt.end())); return result;}// Driver Codeint main(){ vector<int> val{ 60, 100, 120 }; vector<int> wt{ 10, 20, 30 }; int W = 50; int n = val.size(); int maximum = knapSack(W, wt, val, n); cout << "Maximum Profit = "; cout << maximum; return 0;} |
Java
// Java implementation to print all// the possible solutions of the// 0/1 Knapsack problemimport java.util.*;public class Main { // Utility function to find the maximum of the two // elements static int max(int a, int b) { return (a > b) ? a : b; } // Function to find the all the possible solutions of // the 0/1 knapSack problem static int knapSack(int W, List<Integer> wt, List<Integer> val, int n) { // Mapping weights with Profits Map<Integer, Integer> umap = new HashMap<>(); Set<List<Map.Entry<Integer, Integer> > > setSol = new HashSet<>(); // Making Pairs and inserting into the map for (int i = 0; i < n; i++) { umap.put(wt.get(i), val.get(i)); } int result = Integer.MIN_VALUE; int remaining_weight; int sum = 0; // Loop to iterate over all the possible // permutations of array do { sum = 0; // Initially bag will be empty remaining_weight = W; List<Map.Entry<Integer, Integer> > possible = new ArrayList<>(); // Loop to fill up the bag until there is no // weight such which is less than remaining // weight of the 0-1 knapSack for (int i = 0; i < n; i++) { if (wt.get(i) <= remaining_weight) { remaining_weight -= wt.get(i); Integer valAtWtI = umap.get(wt.get(i)); sum += valAtWtI; possible.add( new AbstractMap.SimpleEntry<>( wt.get(i), valAtWtI)); } } Collections.sort( possible, Comparator.comparingInt(Map.Entry::getKey)); if (sum > result) { result = sum; } if (!setSol.contains(possible)) { for (Map.Entry<Integer, Integer> sol : possible) { System.out.print(sol.getKey() + ": " + sol.getValue() + ", "); } System.out.println(); setSol.add(possible); } } while (nextPermutation(wt)); return result; } // Utility function to generate the next permutation static boolean nextPermutation(List<Integer> arr) { int i = arr.size() - 2; while (i >= 0 && arr.get(i) >= arr.get(i + 1)) { i--; } if (i < 0) { return false; } int j = arr.size() - 1; while (arr.get(j) <= arr.get(i)) { j--; } int temp = arr.get(i); arr.set(i, arr.get(j)); arr.set(j, temp); Collections.reverse(arr.subList(i + 1, arr.size())); return true; } // Driver code public static void main(String[] args) { List<Integer> val = new ArrayList<>(Arrays.asList(60, 100, 120)); List<Integer> wt = new ArrayList<>(Arrays.asList(10, 20, 30)); int W = 50; int n = val.size(); int maximum = knapSack(W, wt, val, n); System.out.println("Maximum Profit = " + maximum); }}// This code was contributed by rutikbhosale |
Python3
# Python3 implementation to print all# the possible solutions of the# 0/1 Knapsack problemINT_MIN=-2147483648def nextPermutation(nums: list) -> None: """ Do not return anything, modify nums in-place instead. """ if sorted(nums,reverse=True)==nums: return None n=len(nums) brk_point=-1 for pos in range(n-1,0,-1): if nums[pos]>nums[pos-1]: brk_point=pos break else: nums.sort() return replace_with=-1 for j in range(brk_point,n): if nums[j]>nums[brk_point-1]: replace_with=j else: break nums[replace_with],nums[brk_point-1]=nums[brk_point-1],nums[replace_with] nums[brk_point:]=sorted(nums[brk_point:]) return nums# Function to find the all the# possible solutions of the # 0/1 knapSack problemdef knapSack(W, wt, val, n): # Mapping weights with Profits umap=dict() set_sol=set() # Making Pairs and inserting # o the map for i in range(n) : umap[wt[i]]=val[i] result = INT_MIN remaining_weight=0 sum = 0 # Loop to iterate over all the # possible permutations of array while True: sum = 0 # Initially bag will be empty remaining_weight = W possible=[] # Loop to fill up the bag # until there is no weight # such which is less than # remaining weight of the # 0-1 knapSack for i in range(n) : if (wt[i] <= remaining_weight) : remaining_weight -= wt[i] sum += (umap[wt[i]]) possible.append((wt[i], umap[wt[i]]) ) possible.sort() if (sum > result) : result = sum if (tuple(possible) not in set_sol): for sol in possible: print(sol[0], ": ", sol[1], ", ",end='') print() set_sol.add(tuple(possible)) if not nextPermutation(wt): break return result# Driver Codeif __name__ == '__main__': val=[60, 100, 120] wt=[10, 20, 30] W = 50 n = len(val) maximum = knapSack(W, wt, val, n) print("Maximum Profit =",maximum)#This code was contributed by Amartya Ghosh |
Javascript
// Utility function to find the maximum of the two elementsfunction max(a, b) { return (a > b) ? a : b; }// Function to find the all the possible solutions of the 0/1 knapSack problemfunction knapSack(W, wt, val, n) { // Mapping weights with Profits let umap = new Map(); let set_sol = new Set(); // Making Pairs and inserting into the map for (let i = 0; i < n; i++) { umap.set(wt[i], val[i]); } let result = Number.MIN_SAFE_INTEGER; let remaining_weight, sum; // Loop to iterate over all the possible permutations of array do { sum = 0; // Initially bag will be empty remaining_weight = W; let possible = []; // Loop to fill up the bag until there is no weight such which is less than remaining weight of the 0-1 knapSack for (let i = 0; i < n; i++) { if (wt[i] <= remaining_weight) { remaining_weight -= wt[i]; let val = umap.get(wt[i]); sum += val; possible.push([wt[i], val]); } } possible.sort((a, b) => a[0] - b[0]); if (sum > result) { result = sum; } if (!set_sol.has(JSON.stringify(possible))) { for (let i = 0; i < possible.length; i++) { console.log(possible[i][0] + ": " + possible[i][1] + ", "); } console.log(); set_sol.add(JSON.stringify(possible)); } } while (nextPermutation(wt)); return result;}// Function to generate the next permutation of arrayfunction nextPermutation(a) { let i = a.length - 2; while (i >= 0 && a[i] >= a[i + 1]) { i--; } if (i < 0) { return false; } let j = a.length - 1; while (a[j] <= a[i]) { j--; } let temp = a[i]; a[i] = a[j]; a[j] = temp; for (let l = i + 1, r = a.length - 1; l < r; l++, r--) { temp = a[l]; a[l] = a[r]; a[r] = temp; } return true;}// Driver Codefunction main() { let val = [60, 100, 120]; let wt = [10, 20, 30]; let W = 50; let n = val.length; let maximum = knapSack(W, wt, val, n); console.log("Maximum Profit = " + maximum);}main(); |
C#
using System;using System.Collections.Generic;using System.Linq;class MainClass { // Utility function to find the maximum of the two elements static int max(int a, int b) { return (a > b) ? a : b; } // Function to find the all the possible solutions of // the 0/1 knapSack problem static int knapSack(int W, List<int> wt, List<int> val, int n) { // Mapping weights with Profits Dictionary<int, int> umap = new Dictionary<int, int>(); HashSet<List<KeyValuePair<int, int>>> setSol = new HashSet<List<KeyValuePair<int, int>>>(); // Making Pairs and inserting into the map for (int i = 0; i < n; i++) { umap.Add(wt[i], val[i]); } int result = int.MinValue; int remaining_weight; int sum = 0; // Loop to iterate over all the possible permutations of array do { sum = 0; // Initially bag will be empty remaining_weight = W; List<KeyValuePair<int, int>> possible = new List<KeyValuePair<int, int>>(); // Loop to fill up the bag until there is no // weight such which is less than remaining // weight of the 0-1 knapSack for (int i = 0; i < n; i++) { if (wt[i] <= remaining_weight) { remaining_weight -= wt[i]; int valAtWtI = umap[wt[i]]; sum += valAtWtI; possible.Add( new KeyValuePair<int, int>( wt[i], valAtWtI)); } } possible.Sort( (x, y) => x.Key.CompareTo(y.Key)); if (sum > result) { result = sum; } if (!setSol.Contains(possible)) { foreach (KeyValuePair<int, int> sol in possible) { Console.Write(sol.Key + ": " + sol.Value + ", "); } Console.WriteLine(); setSol.Add(possible); } } while (nextPermutation(wt)); return result; } // Utility function to generate the next permutation static bool nextPermutation(List<int> arr) { int i = arr.Count - 2; while (i >= 0 && arr[i] >= arr[i + 1]) { i--; } if (i < 0) { return false; } int j = arr.Count - 1; while (arr[j] <= arr[i]) { j--; } int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; arr.Reverse(i + 1, arr.Count - i - 1); return true; } // Driver code public static void Main(string[] args) { List<int> val = new List<int>{60, 100, 120}; List<int> wt = new List<int>{10, 20, 30}; int W = 50; int n = val.Count; int maximum = knapSack(W, wt, val, n); Console.WriteLine("Maximum Profit = " + maximum); }} |
Output
10: 60, 20: 100, 10: 60, 30: 120, 20: 100, 30: 120, Maximum Profit = 220
Time complexity : O(N! * N), where N is the number of items. The code uses permutation to generate all possible combinations of items and then performs a search operation to find the optimal solution.
Space complexity : O(N), as the code uses a map and set to store the solution, which has a maximum size of N.
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