# Best Time to Buy and Sell Stock LeetCode Solution Share:

Best Time to Buy and Sell Stock You are given an array `prices` where `prices[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: prices = [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: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.
```

Constraints:

• `1 <= prices.length <= 105`
• `0 <= prices[i] <= 104`

Time: O(n)
Space: O(1)

### C++

``````class Solution {
public:
int maxProfit(vector<int>& prices) {
int sellOne = 0;
int holdOne = INT_MIN;

for (const int price : prices) {
sellOne = max(sellOne, holdOne + price);
holdOne = max(holdOne, -price);
}

return sellOne;
}
};
``````

### Java

``````class Solution {
public int maxProfit(int[] prices) {
int sellOne = 0;
int holdOne = Integer.MIN_VALUE;

for (final int price : prices) {
sellOne = Math.max(sellOne, holdOne + price);
holdOne = Math.max(holdOne, -price);
}

return sellOne;
}
}
``````

### Python

``````class Solution:
def maxProfit(self, prices: List[int]) -> int:
sellOne = 0
holdOne = -math.inf

for price in prices:
sellOne = max(sellOne, holdOne + price)
holdOne = max(holdOne, -price)

return sellOne

``````

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