<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 Memo(int idx,int buy,vector<int>& prices,vector<vector<int>>& dp){
    //     if(idx==prices.size()) return 0;
    //     if(dp[idx][buy]!=-1) return dp[idx][buy];
    //     int profit = 0;
    //     if(buy){
    //         profit = max(-prices[idx]+Memo(idx+1,0,prices,dp),Memo(idx+1,1,prices,dp));
    //     }
    //     else{
    //         profit = max(prices[idx]+Memo(idx+1,1,prices,dp),Memo(idx+1,0,prices,dp));
    //     }
    //     return dp[idx][buy] = profit;
    // }
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        // vector<vector<int>> dp(n+1,vector<int>(2,0));
        vector<int> nxt(2,0);
        for(int i=n-1;i>=0;i--){
            vector<int> cur(2,0);
            cur[1] = max(-prices[i]+nxt[0],nxt[1]);
            cur[0] = max(prices[i]+nxt[1],nxt[0]);
            nxt = cur;
        }
        return nxt[1];
    }
};
17000