You are given an array prices where prices[i] is the price of a given stock on the ith day.

Find the maximum profit you can achieve. You may complete at most two transactions.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

// Example 1:

// Input: prices = [3,3,5,0,0,3,1,4]
// Output: 6
// Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
// Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.
// Example 2:

// Input: prices = [1,2,3,4,5]
// Output: 4
//Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
// Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.

class Solution
{
private:
    int buse(vector<vector<vector<int>>> &dp, vector<int> &prices, int buyorsell, int i, int k)
    {
        if (k == 0)
            return 0;
        if (i == prices.size())
        {
            return 0;
        }
        if (dp[i][buyorsell][k] != -1)
            return dp[i][buyorsell][k];

        if (buyorsell)
        {
            int buy = -prices[i] + buse(dp, prices, 0, i + 1, k);
            int dont_buy = buse(dp, prices, 1, i + 1, k);
            return dp[i][buyorsell][k] = max(buy, dont_buy);
        }
        else
        {
            int sell = prices[i] + buse(dp, prices, 1, i + 1, k - 1);
            int dont_sell = buse(dp, prices, 0, i + 1, k);
            return dp[i][buyorsell][k] = max(sell, dont_sell);
        }
        return dp[i][buyorsell][k];
    }

public:
    int maxProfit(vector<int> &prices)
    {
        int n = prices.size();
        vector<vector<vector<int>>> dp(n, vector<vector<int>>(2, vector<int>(3, -1)));
        return buse(dp, prices, 1, 0, 2);
    }
};