<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>

#pragma GCC optimize("Ofast")
#pragma GCC target("avx2,bmi,bmi2,popcnt,lzcnt")

static const int FAST__ = [](){
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  return 0;
}();

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<vector<vector<int>>> dp(n+1,vector<vector<int>>(2,vector<int>(3,0)));
        for(int idx=n-1;idx>=0;idx--) {
            for(int buy = 0;buy<=1;buy++) {
                for(int remTransactions=1;remTransactions<=2;remTransactions++) {
                    int profit = 0;
                    if(buy) {
                        profit = max(-prices[idx]+dp[idx+1][0][remTransactions],0+dp[idx+1][1][remTransactions]);
                    }
                    else {
                        profit = max(prices[idx]+dp[idx+1][1][remTransactions-1],0+dp[idx+1][0][remTransactions]);
                    }

                    dp[idx][buy][remTransactions] = profit;
                }
            }
        }

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

};
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