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| 1 | +class Solution { |
| 2 | +public: |
| 3 | + // Recursive function to calculate the maximum profit |
| 4 | + // `prices`: vector of stock prices |
| 5 | + // `index`: current day index |
| 6 | + // `transactionNo`: current transaction number (0-based, where even = buy, odd = sell) |
| 7 | + // `k`: maximum number of transactions allowed |
| 8 | + int solve(vector<int>& prices, int index, int transactionNo, int k) { |
| 9 | + // Base case: if we reach the end of the prices array or have completed `2 * k` transactions |
| 10 | + if (index == prices.size() || transactionNo == 2 * k) return 0; |
| 11 | + |
| 12 | + int profit = 0; // Variable to store the profit for the current state |
| 13 | + |
| 14 | + // Check if the current transaction is a "buy" (even transactionNo) |
| 15 | + if (transactionNo % 2 == 0) { |
| 16 | + // Option 1: Buy the stock on the current day and move to the next transaction |
| 17 | + int buyStock = -prices[index] + solve(prices, index + 1, transactionNo + 1, k); |
| 18 | + // Option 2: Skip buying and move to the next day |
| 19 | + int notBuyStock = 0 + solve(prices, index + 1, transactionNo, k); |
| 20 | + // Maximize the profit between buying and skipping |
| 21 | + profit = max(buyStock, notBuyStock); |
| 22 | + } else { |
| 23 | + // If the current transaction is a "sell" (odd transactionNo) |
| 24 | + // Option 1: Sell the stock on the current day and move to the next transaction |
| 25 | + int sellStock = +prices[index] + solve(prices, index + 1, transactionNo + 1, k); |
| 26 | + // Option 2: Skip selling and move to the next day |
| 27 | + int notSellStock = 0 + solve(prices, index + 1, transactionNo, k); |
| 28 | + // Maximize the profit between selling and skipping |
| 29 | + profit = max(sellStock, notSellStock); |
| 30 | + } |
| 31 | + |
| 32 | + return profit; // Return the profit for the current state |
| 33 | + } |
| 34 | + |
| 35 | + // Function to calculate the maximum profit with at most `k` transactions |
| 36 | + // `k`: maximum number of transactions allowed |
| 37 | + // `prices`: vector of stock prices |
| 38 | + int maxProfit(int k, vector<int>& prices) { |
| 39 | + int n = prices.size(); // Total number of days |
| 40 | + |
| 41 | + // Call the recursive function starting from day 0 and transaction number 0 |
| 42 | + return solve(prices, 0, 0, k); |
| 43 | + } |
| 44 | +}; |
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