Given n items where each item has some weight and profit associated with it and also given a bag with capacity W, [i.e., the bag can hold at most W weight in it]. The task is to put the items into the bag such that the sum of profits associated with them is the maximum possible.
Note: The constraint here is we can either put an item completely into the bag or cannot put it at all [It is not possible to put a part of an item into the bag].
Input: W = 4, profit[] = [1, 2, 3], weight[] = [4, 5, 1]
Output: 3
Explanation: There are two items which have weight less than or equal to 4. If we select the item with weight 4, the possible profit is 1. And if we select the item with weight 1, the possible profit is 3. So the maximum possible profit is 3. Note that we cannot put both the items with weight 4 and 1 together as the capacity of the bag is 4.
Input: W = 3, profit[] = [1, 2, 3], weight[] = [4, 5, 6]
Output: 0
[Naive Approach] Using Recursion O(2^n) Time and O(n) Space
A simple solution is to consider all subsets of items and calculate the total weight and value of all subsets. Consider the only subsets whose total weight is smaller than W. From all such subsets, pick the subset with maximum value.
Optimal Substructure: To consider all subsets of items, there can be two cases for every item.
- Case 1: The item is included in the optimal subset.
- Case 2: The item is not included in the optimal set.
Follow the below steps to solve the problem:
The maximum value obtained from ‘n’ items is the max of the following two values.
- Case 1 (pick the nth item): Value of the nth item + maximum value obtained by remaining (n-1) items and remaining weight i.e. (W-weight of the nth item).
- Case 2 (don’t pick the nth item): Maximum value obtained by (n-1) items and W weight.
- If the weight of the ‘nth‘ item is greater than ‘W’, then the nth item cannot be included and Case 2 is the only possibility.
C++ #include <bits/stdc++.h> using namespace std; // Returns the maximum value that // can be put in a knapsack of capacity W int knapsackRec(int W, vector<int> &val, vector<int> &wt, int n) { // Base Case if (n == 0 || W == 0) return 0; int pick = 0; // Pick nth item if it does not exceed the capacity of knapsack if (wt[n - 1] <= W) pick = val[n - 1] + knapsackRec(W - wt[n - 1], val, wt, n - 1); // Don't pick the nth item int notPick = knapsackRec(W, val, wt, n - 1); return max(pick, notPick); } int knapsack(int W, vector<int> &val, vector<int> &wt) { int n = val.size(); return knapsackRec(W, val, wt, n); } int main() { vector<int> val = {1, 2, 3}; vector<int> wt = {4, 5, 1}; int W = 4; cout << knapsack(W, val, wt) << endl; return 0; } import java.util.*; class GfG { // Returns the maximum value that // can be put in a knapsack of capacity W static int knapsackRec(int W, int[] val, int[] wt, int n) { // Base Case if (n == 0 || W == 0) return 0; int pick = 0; // Pick nth item if it does not exceed the capacity of knapsack if (wt[n - 1] <= W) pick = val[n - 1] + knapsackRec(W - wt[n - 1], val, wt, n - 1); // Don't pick the nth item int notPick = knapsackRec(W, val, wt, n - 1); return Math.max(pick, notPick); } static int knapsack(int W, int[] val, int[] wt) { int n = val.length; return knapsackRec(W, val, wt, n); } public static void main(String[] args) { int[] val = {1, 2, 3}; int[] wt = {4, 5, 1}; int W = 4; System.out.println(knapsack(W, val, wt)); } } # Returns the maximum value that # can be put in a knapsack of capacity W def knapsackRec(W, val, wt, n): # Base Case if n == 0 or W == 0: return 0 pick = 0 # Pick nth item if it does not exceed the capacity of knapsack if wt[n - 1] <= W: pick = val[n - 1] + knapsackRec(W - wt[n - 1], val, wt, n - 1) # Don't pick the nth item notPick = knapsackRec(W, val, wt, n - 1) return max(pick, notPick) def knapsack(W, val, wt): n = len(val) return knapsackRec(W, val, wt, n) if __name__ == "__main__": val = [1, 2, 3] wt = [4, 5, 1] W = 4 print(knapsack(W, val, wt))
using System; class GfG { // Returns the maximum value that // can be put in a knapsack of capacity W static int knapsackRec(int W, int[] val, int[] wt, int n) { // Base Case if (n == 0 || W == 0) return 0; int pick = 0; // Pick nth item if it does not exceed the capacity of knapsack if (wt[n - 1] <= W) pick = val[n - 1] + knapsackRec(W - wt[n - 1], val, wt, n - 1); // Don't pick the nth item int notPick = knapsackRec(W, val, wt, n - 1); return Math.Max(pick, notPick); } static int knapsack(int W, int[] val, int[] wt) { int n = val.Length; return knapsackRec(W, val, wt, n); } static void Main() { int[] val = { 1, 2, 3 }; int[] wt = { 4, 5, 1 }; int W = 4; Console.WriteLine(knapsack(W, val, wt)); } } // Returns the maximum value that // can be put in a knapsack of capacity W function knapsackRec(W, val, wt, n) { // Base Case if (n === 0 || W === 0) return 0; let pick = 0; // Pick nth item if it does not exceed the capacity of knapsack if (wt[n - 1] <= W) pick = val[n - 1] + knapsackRec(W - wt[n - 1], val, wt, n - 1); // Don't pick the nth item let notPick = knapsackRec(W, val, wt, n - 1); return Math.max(pick, notPick); } function knapsack(W, val, wt) { let n = val.length; return knapsackRec(W, val, wt, n); } // Driver Code let val = [1, 2, 3]; let wt = [4, 5, 1]; let W = 4; console.log(knapsack(W, val, wt)); Below is an example run of the above implementation.
[Better Approach 1] Using Top-Down DP (Memoization)- O(n x W) Time and Space
Note: The above function using recursion computes the same subproblems again and again. See the following recursion tree, K(1, 1) is being evaluated twice.
As there are repetitions of the same subproblem again and again we can implement the following idea to solve the problem.
If we get a subproblem the first time, we can solve this problem by creating a 2-D array that can store a particular state (n, w). Now if we come across the same state (n, w) again instead of calculating it i again we can directly return its result stored in the table in constant time.
C++ #include <bits/stdc++.h> using namespace std; // Returns the maximum value that // can be put in a knapsack of capacity W int knapsackRec(int W, vector<int> &val, vector<int> &wt, int n, vector<vector<int>> &memo) { // Base Case if (n == 0 || W == 0) return 0; // Check if we have previously calculated the same subproblem if(memo[n][W] != -1) return memo[n][W]; int pick = 0; // Pick nth item if it does not exceed the capacity of knapsack if (wt[n - 1] <= W) pick = val[n - 1] + knapsackRec(W - wt[n - 1], val, wt, n - 1, memo); // Don't pick the nth item int notPick = knapsackRec(W, val, wt, n - 1, memo); // Store the result in memo[n][W] and return it return memo[n][W] = max(pick, notPick); } int knapsack(int W, vector<int> &val, vector<int> &wt) { int n = val.size(); // Memoization table to store the results vector<vector<int>> memo(n + 1, vector<int>(W + 1, -1)); return knapsackRec(W, val, wt, n, memo); } int main() { vector<int> val = {1, 2, 3}; vector<int> wt = {4, 5, 1}; int W = 4; cout << knapsack(W, val, wt) << endl; return 0; } import java.util.*; class GfG { // Returns the maximum value that // can be put in a knapsack of capacity W static int knapsackRec(int W, int[] val, int[] wt, int n, int[][] memo) { // Base Case if (n == 0 || W == 0) return 0; // Check if we have previously calculated the same subproblem if (memo[n][W] != -1) return memo[n][W]; int pick = 0; // Pick nth item if it does not exceed the capacity of knapsack if (wt[n - 1] <= W) pick = val[n - 1] + knapsackRec(W - wt[n - 1], val, wt, n - 1, memo); // Don't pick the nth item int notPick = knapsackRec(W, val, wt, n - 1, memo); // Store the result in memo[n][W] and return it return memo[n][W] = Math.max(pick, notPick); } static int knapsack(int W, int[] val, int[] wt) { int n = val.length; // Memoization table to store the results int[][] memo = new int[n + 1][W + 1]; // Initialize memoization table with -1 for (int i = 0; i <= n; i++) { for (int j = 0; j <= W; j++) memo[i][j] = -1; } return knapsackRec(W, val, wt, n, memo); } public static void main(String[] args) { int[] val = { 1, 2, 3 }; int[] wt = { 4, 5, 1 }; int W = 4; System.out.println(knapsack(W, val, wt)); } } # Returns the maximum value that # can be put in a knapsack of capacity W def knapsackRec(W, val, wt, n, memo): # Base Case if n == 0 or W == 0: return 0 # Check if we have previously calculated the same subproblem if memo[n][W] != -1: return memo[n][W] pick = 0 # Pick nth item if it does not exceed the capacity of knapsack if wt[n - 1] <= W: pick = val[n - 1] + knapsackRec(W - wt[n - 1], val, wt, n - 1, memo) # Don't pick the nth item notPick = knapsackRec(W, val, wt, n - 1, memo) # Store the result in memo[n][W] and return it memo[n][W] = max(pick, notPick) return memo[n][W] def knapsack(W, val, wt): n = len(val) # Memoization table to store the results memo = [[-1] * (W + 1) for _ in range(n + 1)] return knapsackRec(W, val, wt, n, memo) if __name__ == "__main__": val = [1, 2, 3] wt = [4, 5, 1] W = 4 print(knapsack(W, val, wt))
using System; class GfG { // Returns the maximum value that // can be put in a knapsack of capacity W static int KnapsackRec(int W, int[] val, int[] wt, int n, ref int[,] memo) { // Base Case if (n == 0 || W == 0) return 0; // Check if we have previously calculated the same subproblem if (memo[n, W] != -1) return memo[n, W]; int pick = 0; // Pick nth item if it does not exceed the capacity of knapsack if (wt[n - 1] <= W) pick = val[n - 1] + KnapsackRec(W - wt[n - 1], val, wt, n - 1, ref memo); // Don't pick the nth item int notPick = KnapsackRec(W, val, wt, n - 1, ref memo); // Store the result in memo[n, W] and return it return memo[n, W] = Math.Max(pick, notPick); } static int Knapsack(int W, int[] val, int[] wt) { int n = val.Length; // Memoization table to store the results int[,] memo = new int[n + 1, W + 1]; // Initialize memo table with -1 for (int i = 0; i <= n; i++) { for (int j = 0; j <= W; j++) { memo[i, j] = -1; } } return KnapsackRec(W, val, wt, n, ref memo); } static void Main() { int[] val = { 1, 2, 3 }; int[] wt = { 4, 5, 1 }; int W = 4; Console.WriteLine(Knapsack(W, val, wt)); } } // Returns the maximum value that // can be put in a knapsack of capacity W function knapsackRec(W, val, wt, n, memo) { // Base Case if (n === 0 || W === 0) return 0; // Check if we have previously calculated the same subproblem if (memo[n][W] !== -1) return memo[n][W]; let pick = 0; // Pick nth item if it does not exceed the capacity of knapsack if (wt[n - 1] <= W) pick = val[n - 1] + knapsackRec(W - wt[n - 1], val, wt, n - 1, memo); // Don't pick the nth item let notPick = knapsackRec(W, val, wt, n - 1, memo); // Store the result in memo[n][W] and return it memo[n][W] = Math.max(pick, notPick); return memo[n][W]; } function knapsack(W, val, wt) { const n = val.length; // Memoization table to store the results const memo = Array.from({ length: n + 1 }, () => Array(W + 1).fill(-1)); return knapsackRec(W, val, wt, n, memo); } // Driver Code const val = [1, 2, 3]; const wt = [4, 5, 1]; const W = 4; console.log(knapsack(W, val, wt)); [Better Approach 2] Using Bottom-Up DP (Tabulation) – O(n x W) Time and Space
There are two parameters that change in the recursive solution and these parameters go from 0 to n and 0 to W. So we create a 2D dp[][] array of size (n+1) x (W+1), such that dp[i][j] stores the maximum value we can get using i items such that the knapsack capacity is j.
- We first fill the known entries when m is 0 or n is 0.
- Then we fill the remaining entries using the recursive formula.
For each item i and knapsack capacity j, we decide whether to pick the item or not.
- If we don’t pick the item: dp[i][j] remains same as the previous item, that is dp[i – 1][j].
- If we pick the item: dp[i][j] is updated to val[i] + dp[i – 1][j – wt[i]].
C++ #include <bits/stdc++.h> using namespace std; // Returns the maximum value that // can be put in a knapsack of capacity W int knapsack(int W, vector<int> &val, vector<int> &wt) { int n = wt.size(); vector<vector<int>> dp(n + 1, vector<int>(W + 1)); // Build table dp[][] in bottom-up manner for (int i = 0; i <= n; i++) { for (int j = 0; j <= W; j++) { // If there is no item or the knapsack's capacity is 0 if (i == 0 || j == 0) dp[i][j] = 0; else { int pick = 0; // Pick ith item if it does not exceed the capacity of knapsack if(wt[i - 1] <= j) pick = val[i - 1] + dp[i - 1][j - wt[i - 1]]; // Don't pick the ith item int notPick = dp[i - 1][j]; dp[i][j] = max(pick, notPick); } } } return dp[n][W]; } int main() { vector<int> val = {1, 2, 3}; vector<int> wt = {4, 5, 1}; int W = 4; cout << knapsack(W, val, wt) << endl; return 0; } import java.util.*; class GfG { // Returns the maximum value that // can be put in a knapsack of capacity W static int knapsack(int W, int[] val, int[] wt) { int n = wt.length; int[][] dp = new int[n + 1][W + 1]; // Build table dp[][] in bottom-up manner for (int i = 0; i <= n; i++) { for (int j = 0; j <= W; j++) { // If there is no item or the knapsack's capacity is 0 if (i == 0 || j == 0) dp[i][j] = 0; else { int pick = 0; // Pick ith item if it does not exceed the capacity of knapsack if (wt[i - 1] <= j) pick = val[i - 1] + dp[i - 1][j - wt[i - 1]]; // Don't pick the ith item int notPick = dp[i - 1][j]; dp[i][j] = Math.max(pick, notPick); } } } return dp[n][W]; } public static void main(String[] args) { int[] val = {1, 2, 3}; int[] wt = {4, 5, 1}; int W = 4; System.out.println(knapsack(W, val, wt)); } } def knapsack(W, val, wt): n = len(wt) dp = [[0 for _ in range(W + 1)] for _ in range(n + 1)] # Build table dp[][] in bottom-up manner for i in range(n + 1): for j in range(W + 1): # If there is no item or the knapsack's capacity is 0 if i == 0 or j == 0: dp[i][j] = 0 else: pick = 0 # Pick ith item if it does not exceed the capacity of knapsack if wt[i - 1] <= j: pick = val[i - 1] + dp[i - 1][j - wt[i - 1]] # Don't pick the ith item notPick = dp[i - 1][j] dp[i][j] = max(pick, notPick) return dp[n][W] if __name__ == "__main__": val = [1, 2, 3] wt = [4, 5, 1] W = 4 print(knapsack(W, val, wt))
using System; using System.Linq; class GfG { // Returns the maximum value that // can be put in a knapsack of capacity W static int Knapsack(int W, int[] val, int[] wt) { int n = wt.Length; int[,] dp = new int[n + 1, W + 1]; // Build table dp[][] in bottom-up manner for (int i = 0; i <= n; i++) { for (int j = 0; j <= W; j++) { // If there is no item or the knapsack's capacity is 0 if (i == 0 || j == 0) dp[i, j] = 0; else { int pick = 0; // Pick ith item if it does not exceed the capacity of knapsack if (wt[i - 1] <= j) pick = val[i - 1] + dp[i - 1, j - wt[i - 1]]; // Don't pick the ith item int notPick = dp[i - 1, j]; dp[i, j] = Math.Max(pick, notPick); } } } return dp[n, W]; } static void Main() { int[] val = { 1, 2, 3 }; int[] wt = { 4, 5, 1 }; int W = 4; Console.WriteLine(Knapsack(W, val, wt)); } } // Returns the maximum value that // can be put in a knapsack of capacity W function knapsack(W, val, wt) { let n = wt.length; let dp = Array.from({ length: n + 1 }, () => Array(W + 1).fill(0)); // Build table dp[][] in bottom-up manner for (let i = 0; i <= n; i++) { for (let j = 0; j <= W; j++) { // If there is no item or the knapsack's capacity is 0 if (i === 0 || j === 0) dp[i][j] = 0; else { let pick = 0; // Pick ith item if it does not exceed the capacity of knapsack if (wt[i - 1] <= j) pick = val[i - 1] + dp[i - 1][j - wt[i - 1]]; // Don't pick the ith item let notPick = dp[i - 1][j]; dp[i][j] = Math.max(pick, notPick); } } } return dp[n][W]; } // Driver code let val = [1, 2, 3]; let wt = [4, 5, 1]; let W = 4; console.log(knapsack(W, val, wt)); [Expected Approach] Using Bottom-Up DP (Space-Optimized) – O(n x W) Time and O(W) Space
For calculating the current row of the dp[] array we require only previous row, but if we start traversing the rows from right to left then it can be done with a single row only
C++ #include <bits/stdc++.h> using namespace std; // Function to find the maximum profit int knapsack(int W, vector<int> &val, vector<int> &wt) { // Initializing dp vector vector<int> dp(W + 1, 0); // Taking first i elements for (int i = 1; i <= wt.size(); i++) { // Starting from back, so that we also have data of // previous computation of i-1 items for (int j = W; j >= wt[i - 1]; j--) { dp[j] = max(dp[j], dp[j - wt[i - 1]] + val[i - 1]); } } return dp[W]; } int main() { vector<int> val = {1, 2, 3}; vector<int> wt = {4, 5, 1}; int W = 4; cout << knapsack(W, val, wt) << endl; return 0; } import java.util.*; class GfG { // Function to find the maximum profit static int knapsack(int W, int[] val, int[] wt) { // Initializing dp array int[] dp = new int[W + 1]; // Taking first i elements for (int i = 1; i <= wt.length; i++) { // Starting from back, so that we also have data of // previous computation of i-1 items for (int j = W; j >= wt[i - 1]; j--) { dp[j] = Math.max(dp[j], dp[j - wt[i - 1]] + val[i - 1]); } } return dp[W]; } public static void main(String[] args) { int[] val = {1, 2, 3}; int[] wt = {4, 5, 1}; int W = 4; System.out.println(knapsack(W, val, wt)); } } # Function to find the maximum profit def knapsack(W, val, wt): # Initializing dp list dp = [0] * (W + 1) # Taking first i elements for i in range(1, len(wt) + 1): # Starting from back, so that we also have data of # previous computation of i-1 items for j in range(W, wt[i - 1] - 1, -1): dp[j] = max(dp[j], dp[j - wt[i - 1]] + val[i - 1]) return dp[W] if __name__ == "__main__": val = [1, 2, 3] wt = [4, 5, 1] W = 4 print(knapsack(W, val, wt))
using System; class GfG { // Function to find the maximum profit static int Knapsack(int W, int[] val, int[] wt) { // Initializing dp array int[] dp = new int[W + 1]; // Taking first i elements for (int i = 1; i <= wt.Length; i++) { // Starting from back, so that we also have data of // previous computation of i-1 items for (int j = W; j >= wt[i - 1]; j--) { dp[j] = Math.Max(dp[j], dp[j - wt[i - 1]] + val[i - 1]); } } return dp[W]; } static void Main() { int[] val = { 1, 2, 3 }; int[] wt = { 4, 5, 1 }; int W = 4; Console.WriteLine(Knapsack(W, val, wt)); } } // Function to find the maximum profit function knapsack(W, val, wt) { // Initializing dp array let dp = new Array(W + 1).fill(0); // Taking first i elements for (let i = 1; i <= wt.length; i++) { // Starting from back, so that we also have data of // previous computation of i-1 items for (let j = W; j >= wt[i - 1]; j--) { dp[j] = Math.max(dp[j], dp[j - wt[i - 1]] + val[i - 1]); } } return dp[W]; } // Driver Code let val = [1, 2, 3]; let wt = [4, 5, 1]; let W = 4; console.log(knapsack(W, val, wt)); Problems based on 0-1 Knapsack
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