<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][3];
    // int solve(int ind, int buy, int k, vector<int> &a){
    //     if(k==0)return 0;
    //     if(ind == a.size())return 0;
    //     if(dp[ind][buy][k]!=-1)return dp[ind][buy][k];

    //     int nottake = solve(ind+1,buy,k,a);
    //     int take = 0;
    //     if(buy==0){
    //         take = -a[ind] + solve(ind+1,1-buy,k,a);
    //     }else{
    //         take = a[ind] + solve(ind+1,1-buy,k-1,a);
    //     }
    //     return dp[ind][buy][k] = max(take,nottake);
    // }
    int maxProfit(vector<int>& a) {
        // memset(dp,0,sizeof(dp));
        // return solve(0,0,2,a);
        vector<vector<int>> prev(2,vector<int>(3));
        int n = a.size();
        for(int ind = n-1;ind>=0;ind--){
            vector<vector<int>> cur = prev;
            for(int buy = 0; buy<2;buy++){
                for(int k = 1; k<=2;k++){
                    int nottake = prev[buy][k];
                    int take ;
                    if(buy==0){
                        take = -a[ind] + prev[1][k];
                    }else{
                        take = a[ind] + prev[0][k-1];
                    }
                    cur[buy][k] = max(take,nottake);
                }
            }
            prev = cur;
        }
        return prev[0][2];
    }
};
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