<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 {
public:

    int dp[100001][2][2];

    int sol(int i, vector<int>& prices, int buy, int total){
        if(i == prices.size() || total == 2)return 0;
        if(dp[i][buy][total] != -1)return dp[i][buy][total];
        int profit = 0;

        if(buy)profit = max(-prices[i]+sol(i+1, prices, 0, total), sol(i+1, prices, 1, total));
        else profit = max(prices[i]+sol(i+1, prices, 1, total+1), sol(i+1, prices, 0, total));

        return dp[i][buy][total] = profit;
    }

    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        memset(dp, -1, sizeof(dp));

        vector<vector<vector<int>>> dp(n+1, vector<vector<int>>(2, vector<int>(3, 0)));

        for(int i=n-1; i>=0; i--){
            for(int buy=0; buy<=1; buy++){
                for(int total=1; total<=2; total++){
                    int profit = 0;

                    if(buy)profit = max(-prices[i]+dp[i+1][0][total], dp[i+1][1][total]);
                    else profit = max(prices[i]+dp[i+1][1][total-1], dp[i+1][0][total]);

                    dp[i][buy][total] = profit;
                }

            }
        }

        return dp[0][1][2];
    }
};
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