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

Best Time to Buy and Sell Stock II You are given an integer array `prices` where `prices[i]` is the price of a given stock on the `ith` day.

On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day.

Find and return the maximum profit you can achieve.

Example 1:

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

Example 2:

```Input: prices = [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Total profit is 4.
```

Example 3:

```Input: prices = [7,6,4,3,1]
Output: 0
Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.
```

Constraints:

• `1 <= prices.length <= 3 * 104`
• `0 <= prices[i] <= 10`

Time: O(n)
Space: O(1)

### C++

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

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

return sell;
}
};
``````

### Java

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

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

return sell;
}
}
``````

### Python

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

for price in prices:
sell = max(sell, hold + price)
hold = max(hold, sell - price)

return sell

``````

#### Watch Tutorial

Checkout more Solutions here