<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<vector<vector<int>>> dp;
    int profit(vector<int>& prices, int index, bool can_buy, int count){
        if(count == 4 || index == prices.size())
            return 0;
        if(dp[index][can_buy][count] != -1)
            return dp[index][can_buy][count];
        if(can_buy){
            return dp[index][can_buy][count] = max(-prices[index] + profit(prices, index + 1, false, count + 1), profit(prices, index + 1, true, count));
        }else{
            return dp[index][can_buy][count] = max(prices[index] + profit(prices, index + 1, true, count + 1), profit(prices, index + 1, false, count));
        }
    }
    int maxProfit(vector<int>& prices) {
        
        int n = prices.size();
        //dp.resize(n, vector<vector<int>>(2, vector<int>(4, -1)));

        vector<vector<vector<int>>> dp(n + 1 , vector<vector<int>>(2, vector<int>(5, 0)));

        for(int index = n - 1; index >= 0; index--){
            for(int count = 0; count < 4; count++){
                        dp[index][1][count] = max(-prices[index]+dp[index+1][false][count+1], dp[index + 1][true][count]);
                        dp[index][0][count] = max(prices[index] + dp[index + 1][true][count + 1], dp[index+1][false][count]);
            }
        }
        return dp[0][1][0];
        return profit(prices, 0, true, 0);
    }
};
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