<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 maxProfit(vector<int>& prices) {
        int n = prices.size();
        // find the maximum profit by day x on left side and from day x on the right side.
        //find the maximum.
        // calculate the minimums and best trades so far.
        vector<int> minimum_left = {prices[0]};
        vector<int> best_left = {0};
        for (int i = 1; i < n; i++) {
            minimum_left.push_back(min(minimum_left[i-1], prices[i]));
            best_left.push_back(max(best_left[i-1], prices[i] - minimum_left[i]));
        }

        vector<int> best_right(n, 0);
        vector<int> maximum_right(n, 0);
        best_right.back() = 0;
        maximum_right.back() = prices.back();

        for (int i = n-2; i >= 0; i--) {
            maximum_right[i] = max(maximum_right[i+1], prices[i]);
            best_right[i] = max(best_right[i+1], maximum_right[i] - prices[i]);
        }


        // find the best combination:
        int ret = 0;
        for (int i = 0; i < n; i++) {
            // cout << best_left[i] << " " << best_right[i] << endl;
            ret = max(ret, best_right[i] + best_left[i]);
        }
        return ret;

    }
};
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