<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>You want to maximize your profit by choosing a <strong>single day</strong> to buy one stock and choosing a <strong>different day in the future</strong> to sell that stock.</p>

<p>Return <em>the maximum profit you can achieve from this transaction</em>. If you cannot achieve any profit, return <code>0</code>.</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> 5
<strong>Explanation:</strong> Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> prices = [7,6,4,3,1]
<strong>Output:</strong> 0
<strong>Explanation:</strong> In this case, no transactions are done and the 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>4</sup></code></li>
</ul>

class Solution {
public:
    
    void maxToItsRight(int arr[], int n, vector<int>& prices){
        arr[n-1] = prices[n-1] ;
        for(int i = n-2 ; i >=0 ; i--){
            arr[i] = max(arr[i+1], prices[i]);
        }
    }
    
    void minToItsLeft(int arr[], int n, vector<int>& prices){
        arr[0] = prices[0] ;
        for(int i = 1; i< n ; i++){
            arr[i] = min(arr[i-1], prices[i]) ;
        }
    }

    int maxProfit(vector<int>& prices) {
        int n = prices.size() ;
        if (n==1){
            return 0;
        }
        int m2Right[n+1] ; 
        int min2Left[n+1] ;
        maxToItsRight(m2Right, n, prices);
        minToItsLeft(min2Left, n, prices) ;
        int ans=0 ;
        for(int i = 0 ; i < n ;i++){
            ans = max(ans, m2Right[i] - min2Left[i]) ;
        }
        return ans;
    }
};
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