<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>& a) {
        int n=a.size();
        // vector<vector<int>>dp(n+1,vector<int>(5,0));
        vector<int>prev(5,0);
        // return f(0,0,a,n,dp);
        for(int ind=n-1;ind>=0;ind--){
            vector<int>curr(5,0);
            for(int tra=0;tra<4;tra++){
                int profit;
                if(tra%2==0){
                    profit=max(prev[tra+1]-a[ind],prev[tra]);
                }
                else{
                    profit=max(prev[tra+1]+a[ind],prev[tra]);
                }

                curr[tra]=profit;
            }
            prev=curr;
        }
        return prev[0];
    }
    int f(int ind, int tra, vector<int>& a, int n, vector<vector<int>>& dp){
        if(ind==n || tra==4)return 0;
        if(dp[ind][tra]!=-1)return dp[ind][tra];
        int profit;
        if(tra%2==0){
            profit=max(f(ind+1,tra+1,a,n,dp)-a[ind],f(ind+1,tra,a,n,dp));
        }
        else{
            profit=max(f(ind+1,tra+1,a,n,dp)+a[ind],f(ind+1,tra,a,n,dp));
        }

        return dp[ind][tra]=profit;
    }
};
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