|
| 1 | +class Solution { |
| 2 | + public: |
| 3 | + // Recursive function with memoization to solve the knapsack problem |
| 4 | + int solve(int capacity, vector<int> &value, vector<int> &weight, int index, vector<vector<int>>& dp){ |
| 5 | + |
| 6 | + // Base case: If we are at the first item (index 0) |
| 7 | + if(index == 0){ |
| 8 | + // If the weight of the first item is less than or equal to the remaining capacity, return its value |
| 9 | + if(weight[0] <= capacity) return value[0]; |
| 10 | + // Otherwise, return 0 because we cannot include this item in the knapsack |
| 11 | + else return 0; |
| 12 | + } |
| 13 | + |
| 14 | + // If the solution for the current subproblem (index and capacity) has already been computed, return the result from dp table |
| 15 | + if(dp[index][capacity] != -1) return dp[index][capacity]; |
| 16 | + |
| 17 | + // Variable to store the value if we include the current item |
| 18 | + int include = 0; |
| 19 | + // Check if the current item's weight is less than or equal to the remaining capacity |
| 20 | + if(weight[index] <= capacity) { |
| 21 | + // If we include the current item, reduce the capacity by the item's weight, and add its value |
| 22 | + // Recursively solve the remaining problem (remaining capacity and items) |
| 23 | + include = solve(capacity - weight[index], value, weight, index-1, dp) + value[index]; |
| 24 | + } |
| 25 | + |
| 26 | + // Variable to store the value if we exclude the current item |
| 27 | + int exclude = 0; |
| 28 | + // If we exclude the current item, just solve the problem with the remaining items and the same capacity |
| 29 | + exclude = solve(capacity, value, weight, index-1, dp); |
| 30 | + |
| 31 | + // Store the result in the dp table and return the maximum of the two values: including or excluding the item |
| 32 | + dp[index][capacity] = max(include, exclude); |
| 33 | + |
| 34 | + return dp[index][capacity]; |
| 35 | + } |
| 36 | + |
| 37 | + // Function to initialize and call the recursive function to solve the knapsack problem |
| 38 | + int knapSack(int capacity, vector<int> &val, vector<int> &wt) { |
| 39 | + |
| 40 | + int n = wt.size(); // Get the number of items |
| 41 | + |
| 42 | + // Create a memoization table (dp) to store solutions of subproblems |
| 43 | + // dp[i][j] will store the maximum value that can be obtained with the first i items and a knapsack capacity of j |
| 44 | + vector<vector<int>> dp(n+1, vector<int>(capacity + 1, -1)); |
| 45 | + |
| 46 | + // Start the recursive function with the full capacity and all items (index n-1) |
| 47 | + return solve(capacity, val, wt, n-1, dp); |
| 48 | + } |
| 49 | +}; |
0 commit comments