**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 `i`

day.^{th}

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:7Explanation: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:4Explanation: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:0Explanation: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 * 10`

^{4}`0 <= prices[i] <= 10`

### Best Time to Buy and Sell Stock II Solutions

✅**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**