<p>You are given an integer 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>On each day, you may decide to buy and/or sell the stock. You can only hold <strong>at most one</strong> share of the stock at any time. However, you can buy it then immediately sell it on the <strong>same day</strong>.</p>

<p>Find and return <em>the <strong>maximum</strong> profit you can achieve</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> prices = [7,1,5,3,6,4]
<strong>Output:</strong> 7
<strong>Explanation:</strong> Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
Total profit is 4 + 3 = 7.
</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.
Total profit is 4.
</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> There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= prices.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= prices[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:

    int solve(vector<int>&prices, int in,int b,int n,vector<vector<int>>&dp)
    {
        if(in==n)
        {
            return 0;
        }

        if(dp[in][b]!=-1)
        {
            return dp[in][b];
        }

        int profit=0;

        if(b)
        {
            int p1 = prices[in]+ solve(prices,in+1,0,n,dp);
            int p2 = solve(prices,in+1,b,n,dp);

            profit = max(p1,p2);
        }

        else
        {
            int p1 = -prices[in] + solve(prices,in+1,1,n,dp);
            int p2 = solve(prices,in+1,b,n,dp);


            profit = max(p1,p2);

            
        }
        return dp[in][b] = profit;
    }


    int maxProfit(vector<int>& prices) {
        
        int n = prices.size();
        vector<vector<int>>dp(n+1,vector<int>(2,-1));

        int maxi=0;

        dp[n][0]=0;
        dp[n][1]=0;

        for(int i=n-1;i>=0;i--)
        {
            for(int b=0;b<2;b++)
            {
                int profit=0;

                if(b)
                {
                    profit= max(-prices[i]+dp[i+1][0],dp[i+1][1]);
                }
                else
                {
                    profit = max(prices[i]+dp[i+1][1],dp[i+1][0]);
                }

                dp[i][b]=profit;        

            }
           
        }
        return dp[0][1];

        
     
    }
};
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