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| 1 | +class Solution { |
| 2 | +public: |
| 3 | + // Recursive helper function to calculate the maximum profit |
| 4 | + // `prices`: vector of stock prices |
| 5 | + // `index`: current day index in the prices array |
| 6 | + // `buy`: indicates whether we are allowed to buy (1) or need to sell (0) |
| 7 | + // `k`: remaining transactions we are allowed to perform |
| 8 | + int solve(vector<int>& prices, int index, int buy, int k) { |
| 9 | + int n = prices.size(); // Number of days in the prices array |
| 10 | + |
| 11 | + // Base case: if we've reached the end of the array or have no transactions left, return 0 |
| 12 | + if(index == n || k == 0) return 0; |
| 13 | + |
| 14 | + int profit = 0; // Variable to store the maximum profit at the current state |
| 15 | + |
| 16 | + if(buy) { |
| 17 | + // If we are allowed to buy, we have two choices: |
| 18 | + // 1. Buy the stock on this day and subtract its price from profit, then move to the next day with `buy = 0`. |
| 19 | + // 2. Skip buying and move to the next day with `buy = 1`. |
| 20 | + profit = max( |
| 21 | + -prices[index] + solve(prices, index + 1, 0, k), // Buy |
| 22 | + 0 + solve(prices, index + 1, 1, k) // Skip |
| 23 | + ); |
| 24 | + } else { |
| 25 | + // If we are not allowed to buy (we need to sell), we have two choices: |
| 26 | + // 1. Sell the stock on this day and add its price to profit, then move to the next day with `buy = 1` and decrement transactions. |
| 27 | + // 2. Skip selling and move to the next day with `buy = 0`. |
| 28 | + profit = max( |
| 29 | + prices[index] + solve(prices, index + 1, 1, k - 1), // Sell |
| 30 | + 0 + solve(prices, index + 1, 0, k) // Skip |
| 31 | + ); |
| 32 | + } |
| 33 | + |
| 34 | + return profit; // Return the maximum profit for this state |
| 35 | + } |
| 36 | + |
| 37 | + // Main function to calculate the maximum profit |
| 38 | + // `k`: maximum number of transactions allowed |
| 39 | + // `prices`: vector of stock prices |
| 40 | + int maxProfit(int k, vector<int>& prices) { |
| 41 | + // Start the recursive process from day 0, with the ability to buy and `k` transactions remaining |
| 42 | + return solve(prices, 0, 1, k); |
| 43 | + } |
| 44 | +}; |
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