<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 f(int i, int b, vector<int>&p){
    //     // base case
    //     if(i == p.size()) return 0;
    //     // check if u hold already hold a share
    //     if(b==0){
    //         // max(do nothing, buy)
    //         return max(f(i+1,0,p),f(i+1,1,p)-p[i]);
    //     }
    //     else{
    //         // max(do nothing, sell)
    //         return max(f(i+1,1,p),f(i+1,0,p)+p[i]);
    //     }
    // }

    // int f(int i, int b, vector<int>&p, vector<vector<int>> &dp){
    //     // base case
    //     if(i == p.size()) return 0;
    //     // memoization
    //     if(dp[i][b] != -1) return dp[i][b];
    //     // check if u hold already hold a share
    //     if(b==0){
    //         // max(do nothing, buy)
    //         return max(f(i+1,0,p,dp),f(i+1,1,p,dp)-p[i]);
    //     }
    //     else{
    //         // max(do nothing, sell)
    //         return max(f(i+1,1,p,dp),f(i+1,0,p,dp)+p[i]);
    //     }
    // }

    int maxProfit(vector<int>& prices) {
        // return f(0,0,prices);
        int n = prices.size();
        // vector<vector<int>> dp(n,vector<int> (2,-1));
        // return f(0,0,prices,dp);
        
        vector<vector<int>> dp(n+1,vector<int> (2,-1));
        dp[n][0] = 0;
        dp[n][1] = 0;
        for(int i = n-1; i >= 0; i--){
            // can buy then max(do nothing, buy)
            dp[i][0] = max(dp[i+1][0], dp[i+1][1]-prices[i]);
            // can't buy then max(do nothing, sell)
            dp[i][1] = max(dp[i+1][1],dp[i+1][0]+prices[i]);
        }
        //return the case when starting from 0th day and can buy
        return dp[0][0];
    }
};
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