<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 solve(vector<int> &v, int i, int buy,vector<vector<vector<int>>> &dp,int limit)
    {
        if(i>=v.size()) return 0;
        if(limit==0) return 0;

        if(dp[i][buy][limit]!=-1) return dp[i][buy][limit];
        int profit = 0;
        
        if(buy)
        {
            int profit1 = solve(v,i+1,0,dp,limit-1) - v[i];
            int profit2 = solve(v,i+1,1,dp,limit)+0;

            profit = max(profit,max(profit1,profit2));
        }

        else
        {
            int profit1 = solve(v,i+1,1,dp,limit-1)+v[i];
            int profit2 = solve(v,i+1,0,dp,limit);

            profit = max(profit,max(profit1,profit2));
        }

        return dp[i][buy][limit] = profit;
    }

    int maxProfit(vector<int>& prices)
    {
        vector<vector<vector<int>>> dp(prices.size(),vector<vector<int>>(2,vector<int>(5,-1)));
        return solve(prices,0,1,dp,4);
    }

};
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