<p>You are given an array <code>prices</code> where <code>prices[i]</code> is the price of a given stock on the <code>i<sup>th</sup></code> day.</p>

<p>Find the maximum profit you can achieve. You may complete <strong>at most two transactions</strong>.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> prices = [3,3,5,0,0,3,1,4]
<strong>Output:</strong> 6
<strong>Explanation:</strong> 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.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> prices = [1,2,3,4,5]
<strong>Output:</strong> 4
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> prices = [7,6,4,3,1]
<strong>Output:</strong> 0
<strong>Explanation:</strong> In this case, no transaction is done, i.e. max profit = 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= prices.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= prices[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
    // int solve(vector<int>& prices, int i, int f, int k, vector<vector<vector<int>>>& dp) {
    //     if (i >= prices.size() || k >= 2) return 0;

    //     if (dp[k][f][i] != -1) return dp[k][f][i];

    //     int buy = 0, notBuy = 0, sell = 0, notSell = 0;
    //     if (f) {
    //         buy = solve(prices, i+1, 0, k, dp) - prices[i];
    //         notBuy = solve(prices, i+1, 1, k, dp);
    //     } else {
    //         sell = solve(prices, i+1, 1, k+1, dp) + prices[i];
    //         notSell = solve(prices, i+1, 0, k, dp);
    //     }

    //     return dp[k][f][i] = max(max(buy, notBuy), max(sell, notSell));
    // }
public:
    int maxProfit(vector<int>& prices) {
        // vector<vector<vector<int>>> dp(2, vector<vector<int>>(2, vector<int>(prices.size()+1, -1)));
        // return solve(prices, 0, 1, 0, dp);

        int n = prices.size();
        vector<vector<vector<int>>> dp(3, vector<vector<int>>(2, vector<int>(n+1, 0)));

        for (int i = n-1; i > -1; i--) {
            for (int f = 0; f < 2; f++) {
                for (int k = 1; k > -1; k--) {

                    int buy = 0, notBuy = 0, sell = 0, notSell = 0;
                    if (f) {
                        buy = dp[k][0][i+1] - prices[i];
                        notBuy = dp[k][1][i+1];
                    } else {
                        sell = dp[k+1][1][i+1] + prices[i];
                        notSell = dp[k][0][i+1];
                    }

                    dp[k][f][i] = max(max(buy, notBuy), max(sell, notSell));
                }
            }
        }

        return dp[0][1][0];
    }
};
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