<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 Solution1{
public:
    int solve(int k,int count ,vector<int>& prices){
        if(count==0)return 0;
        if(k>=prices.size()-1){
            return 0;
        }
        else{
            int ans=0;
            for(int i=k;i<prices.size();i++){
                for(int j=i+1;j<prices.size();j++){
                    ans=max(ans,prices[j]-prices[i] + solve(j+1,count-1,prices));
                }
            }
            return ans;
        }
    }
    int maxProfit(vector<int>& prices) {
        return solve(0,2,prices);
    }
};
class Solution2 {
public:
    int solve(int k,int count ,vector<int>& prices,vector<vector<int>>&dp){
        if(count==0)return 0;
        if(k>=prices.size()-1){
            return 0;
        }
        if(dp[k][count]!=-1)return dp[k][count];
        else{
            int ans=0;
            for(int i=k;i<prices.size();i++){
                for(int j=i+1;j<prices.size();j++){
                    if(prices[j]>prices[i]){
                        ans=max(ans,prices[j]-prices[i] + solve(j+1,count-1,prices,dp));
                    }
                }
            }
            return dp[k][count]=ans;
        }
    }
    int maxProfit(vector<int>& prices) {
        vector<vector<int>>dp(prices.size()+1,vector<int>(3,-1));
        return solve(0,2,prices,dp);
    }
};
class Solution {
public:
    int sell(int i,int count,vector<int>& prices,vector<vector<int>>&dp1,vector<vector<int>>&dp2){
        if(i>=prices.size()){
            return 0;
        }
        if(count<=0){
            return 0;
        }
        if(dp1[i][count]!=-1){
            return dp1[i][count];
        }
        else{
            int a=prices[i]+buy(i+1,count-1,prices,dp1,dp2);
            int b=sell(i+1,count,prices,dp1,dp2);
            return dp1[i][count]=max(a,b);
        }
    }
    int buy(int i,int count ,vector<int>& prices,vector<vector<int>>&dp1,vector<vector<int>>&dp2){
        if(i>=prices.size()-1){
            return 0;
        }
        if(count<=0){
            return 0;
        }
        if(dp2[i][count]!=-1){
            return dp2[i][count];
        }
        else{
            int a=buy(i+1,count,prices,dp1,dp2);
            int b=-prices[i]+sell(i+1,count,prices,dp1,dp2);
            return dp2[i][count]=max(a,b);
        }
    }
    int maxProfit(vector<int>& prices) {
        vector<vector<int>>dp1(prices.size()+1,(vector<int>(3,-1)));
        vector<vector<int>>dp2(prices.size()+1,(vector<int>(3,-1)));
        return buy(0,2,prices,dp1,dp2);
    }
};
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