<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 f(int ind, int buy,int cap, vector<int>& p,vector<vector<vector<long long int>>> & dp){
    //     if(ind==p.size() || cap==0) return 0 ;
    //     long long int profit;
    //     if(dp[ind][buy][cap]!=-1) return dp[ind][buy][cap];
    //     if(buy){
    //         profit = max(-p[ind]+f(ind+1,0,cap,p,dp) , 0+ f(ind+1,1,cap,p,dp));
    //     }else {
    //         profit = max(p[ind]+f(ind+1,1,cap-1,p,dp),0+f(ind+1,0,cap,p,dp));
    //     }
    //     return dp[ind][buy][cap] = profit;

    // }
    int maxProfit(vector<int>& p) {
        int n = p.size();
        // vector<vector<vector<long long int>>> dp(n+1,
        vector<vector<long long int>> dp(n+1,vector<long long int> (5,0));

        // for(int i=0; i<n; i++){
        //     dp[0][0]=0;
        //     dp[1][0]=0;
        // }
        // for(int i=0; i<3; i++){
        //     dp[n][0][0]=0;
        //     dp[n][1][0]=0;
        // }
        for(int ind = n-1 ; ind >=0 ; ind-- ){
            for(int trans = 3; trans>=0; trans--){
                if(trans%2==0){
                    dp[ind][trans] = max(-p[ind]+dp[ind+1][trans+1], 0+ dp[ind+1][trans]);
                }else {
                    dp[ind][trans] = max(p[ind]+dp[ind+1][trans+1],0+dp[ind+1][trans]);
                }
            }
        }

        return dp[0][0];
    }
};
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