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| 1 | +class Solution { |
| 2 | +public: |
| 3 | + // Helper function to calculate the maximum profit |
| 4 | + int solve(vector<int>& prices, int index, int buy, int fee) { |
| 5 | + int n = prices.size(); // Get the size of the prices array |
| 6 | + |
| 7 | + // Base case: If we've reached the end of the array, no more transactions can be made |
| 8 | + if(index == n) return 0; |
| 9 | + |
| 10 | + int profit = 0; |
| 11 | + |
| 12 | + // If we are in a buying state (buy == 1), we can either: |
| 13 | + // 1. Buy the stock at current price (this is represented by -prices[index]) |
| 14 | + // and move to the next index (index+1) in a "sell" state (buy = 0). |
| 15 | + // 2. Skip this index (do nothing) and stay in the "buy" state (buy = 1). |
| 16 | + if(buy) { |
| 17 | + profit = max((-prices[index] + solve(prices, index+1, 0, fee)), // Buy and move to the sell state |
| 18 | + (0 + solve(prices, index+1, 1, fee))); // Do nothing and stay in the buy state |
| 19 | + } else { |
| 20 | + // If we are in a selling state (buy == 0), we can either: |
| 21 | + // 1. Sell the stock at current price (this is represented by +prices[index] - fee) |
| 22 | + // and move to the next index (index+1) in a "buy" state (buy = 1). |
| 23 | + // 2. Skip this index (do nothing) and stay in the "sell" state (buy = 0). |
| 24 | + profit = max((+prices[index] - fee + solve(prices, index+1, 1, fee)), // Sell and move to the buy state |
| 25 | + (0 + solve(prices, index+1, 0, fee))); // Do nothing and stay in the sell state |
| 26 | + } |
| 27 | + |
| 28 | + // Return the maximum profit for the current state |
| 29 | + return profit; |
| 30 | + } |
| 31 | + |
| 32 | + // Main function to calculate the maximum profit with transaction fee |
| 33 | + int maxProfit(vector<int>& prices, int fee) { |
| 34 | + // Start solving from index 0 with the possibility of buying (buy = 1) |
| 35 | + return solve(prices, 0, 1, fee); |
| 36 | + } |
| 37 | +}; |
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