You are given an array a where a[i] is the price of a given stock on the ith day.

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

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return `0`.

Example 1:

```Input: a = [7,1,5,3,6,4]
Output: 5
Explanation: 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.
```

Example 2:

```Input: a = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.
```

method: greedy approach

1)We need to find the largest price following the smallest price.
2)We can maintain two variables - s and c corresponding to the smallest valley and maximum profit.
3)now traverse the array and  keep the track of maximum difference between selling price (s) and minprice(c) obtained so far respectively and keep updating the values of s and c.
4)return c at the end

c++ implementation:

```int maxProfit(vector<int>& a)
{
int s=a;int c=0;
for(int i=1;i<a.size();i++)
{
if(a[i]-s> c)
{
c=a[i]-s;
}
else
{
if(a[i]<s)
s=a[i];
}
}
return c;

}```

Time Complexity: O(n)
space Complexity: O(1)

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