<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:
    vector<int> func(vector<int>&nums){
        int n = nums.size();
        vector<int>ans;
        int mini = INT_MAX, maxi = INT_MIN;
        for(int i=0;i<n;i++){
          if(nums[i] < mini) mini = nums[i];
          ans.push_back(nums[i] - mini);
        }
        for(int i=1;i<n;i++){
            ans[i] = max(ans[i], ans[i-1]);
        }
        return ans;
    }
    vector<int> func1(vector<int>&nums){
        int n = nums.size();
        vector<int>ans(n);
        int maxi = INT_MIN;
        for(int i=n-1;i>=0;i--){
            if(maxi<nums[i]){
                maxi = nums[i];
            }
            ans[i] = abs(nums[i] - maxi);
        }
        // for(auto x: ans) cout<<x<<" ";
        for(int i=n-2;i>=0;i--){
            ans[i] = max(ans[i], ans[i+1]);
        }
        return ans;
    }
    int maxProfit(vector<int>& nums) {
        int n = nums.size();
        vector<int>maxi1(n), maxi2(n);
        maxi1 = func(nums);
        maxi2 = func1(nums);
        int ans = 0;
        for(int i=0;i<n;i++){
            ans = max(ans, maxi1[i] + maxi2[i]);
        }
        return ans;
    }
};
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