<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:
    void printVec(vector<int> vec) {
        for(int i=0;i<size(vec);i++)
            cout << vec[i] << " ";
        cout << endl;
    }
    int maxProfit(vector<int>& prices) {
        vector<int> left, right;
        int mx=0,mn=prices[0],n=size(prices),cur=0;
        for(int i=0;i<n;i++) {
            if(prices[i] > mx)
                mx = prices[i];
            
            if(prices[i] < mn) {
                mn = prices[i];
                mx = prices[i];
            }
            cur = max(mx-mn, cur);
            left.push_back(cur);
        }
        mx=0,mn=prices[n-1],cur=0;
        for(int i=n-1;i>=0;i--) {
            if(prices[i] < mn)
                mn = prices[i];
            
            if(prices[i] > mx) {
                mn = prices[i];
                mx = prices[i];
            }
            cur = max(mx-mn, cur);
            right.push_back(cur);
        }
        reverse(right.begin(), right.end());
        printVec(left);
        printVec(right);

        int ans=right[0];
        for(int i=0;i<n-1;i++) {
            ans = max(ans, left[i]+right[i+1]);
        }
        return max(ans, left[n-1]);
    }
};
100936