<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>&prices, int day, int transactionsLeft, vector<vector<int>>& Memo){
        if(day == prices.size())return 0;
        if(transactionsLeft == 0)return 0;
        
        int& ans = Memo[day][transactionsLeft];
        // cout<<ans<<endl;
        if (ans != -1) {
            // cout<<"memory hit";
            return ans;
        }
        
        // choice 1: no transaction today
        int ans1 = solve(prices, day + 1, transactionsLeft, Memo);
        // choice 2: doing the possible transaction today     
        int ans2 = 0;
        bool buy = (transactionsLeft % 2 == 0);
        ans2 = (buy?-1:1)*prices[day] + solve(prices, day + 1, transactionsLeft - 1, Memo);
        return ans = std::max(ans1, ans2);
    }
    int maxProfit(vector<int>& prices) {
        vector<vector<int>> Memo(prices.size(), vector<int>(5, -1));
        // for (auto& i: Memo) {
        //     for (auto& j: i) {
        //         cout<<j;
        //     }
        // }
        int ans = solve(prices, 0, 4, Memo); // day 0 and max 4 transactions can be done
        return ans;
    }
};
// class Solution {
// public:
    
//     int solve(vector<int>&prices, int day, int transactionsLeft, vector<vector<int>> &Memo){
        
//         if(day == prices.size()){
//             return 0;
//         }
        
//         if(transactionsLeft == 0){
//             return 0;
//         }
        
//         int &ans = Memo[day][transactionsLeft]; 
        
//         if(ans != -1){ // if problem has already been solved 
//             return ans;
//         }
        
//         // choice 1
//         // no transaction today
//         int ans1 = solve(prices, day + 1, transactionsLeft, Memo);
        
        
//         // choice 2
//         // doing the possible transaction today     
//         int ans2 = 0;
//         bool buy = (transactionsLeft % 2 == 0);
        
//         if(buy == true){ // buy
//             ans2 = -prices[day] + solve(prices, day + 1, transactionsLeft - 1, Memo);
//         }else{ // sell
//             ans2 = prices[day] + solve(prices, day + 1, transactionsLeft - 1, Memo);
//         }
//         return ans = max(ans1, ans2); // store ans in memo before returning   
//     }
//     int maxProfit(vector<int>& prices) {
//         vector<vector<int>> Memo(prices.size(), vector<int>(5, -1));
//         int ans = solve(prices, 0, 4, Memo);
//         return ans;
        
//     }
// };
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