30 lines to solve MaxProfit problem

MaxProfit.java

public int maxProfit(int[] prices, int k, int day, int fee) {
    int n = prices.length;
    if (n == 0 || n == 1 || k == 0) {
        return 0;
    }

    int[] buy = new int[n];
    int[] sell = new int[n];
    int[] oldSell = new int[n];
    int maxProfit = 0;

    for (int index = 1; index <= k; index++) {
        buy[0] = -prices[0];
        for (int i = 1; i < n; i++) {
            if (i > day) {
                buy[i]=(Math.max(buy[i - 1], oldSell[i - day - 1] - prices[i]));
            } else {
                buy[i]=(Math.max(buy[i - 1], -prices[i]));
            }
            sell[i]=(Math.max(buy[i - 1] + prices[i] - fee, sell[i - 1]));
        }
        maxProfit = Math.max(maxProfit, sell[n - 1]);

        oldSell = sell.clone();
        buy = new int[n];
        sell = new int[n];
    }
  
    return maxProfit;
}

MaxProfit.swift

func maxProfit(_ prices: [Int], _ k: Int, _ day: Int, _ fee: Int) -> Int {
    let n = prices.count
    if n == 0 || n == 1 || k == 0 {
        return 0
    }
    
    var buy: [Int] = []
    var sell: [Int] = [0]
    var oldSell: [Int] = [Int](repeating: 0, count: n)
    var maxProfit = 0
    
    for var index in 1...k {
        buy.append(-prices[0])
        for var i in 1..<n {
            if i > day {
                buy.append(max(buy[i - 1], oldSell[i - day - 1] - prices[i]))
            } else {
                buy.append(max(buy[i - 1], -prices[i]))
            }
            sell.append(max(buy[i - 1] + prices[i] - fee, sell[i - 1]))
        }
        maxProfit = max(maxProfit, sell[n - 1])

        oldSell = sell
        buy = []
        sell = [0]
    }
    
    return maxProfit
}

Series of stock problems in LeetCode:

1. 121. Best Time to Buy and Sell Stock: maxProfit(prices, 1, 0, 0)
2. 122. Best Time to Buy and Sell Stock II: maxProfit(prices, prices.length / 2, 0, 0)
3. 123. Best Time to Buy and Sell Stock III: maxProfit(prices, 2, 0, 0)
4. 188. Best Time to Buy and Sell Stock IV: maxProfit(prices, Math.abs(k, prices.length / 2, 0, 0)
5. 309. Best Time to Buy and Sell Stock with Cooldown: maxProfit(prices, prices.length / 2, 1, 0)
6. 714. Best Time to Buy and Sell Stock with Transaction Fee: maxProfit(prices, prices.length / 2, 0, fee)