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

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

Find the maximum profit you can achieve. You may complete at most two transactions.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

Example 1:

```Input: prices = [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.```

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.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.
```

Example 3:

```Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
```

Constraints:

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

Time: O(n)
Space: O(1)

### C++

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

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

return sellTwo;
}
};
``````

### Java

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

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

return sellTwo;
}
}
``````

### Python

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

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

return sellTwo
``````

#### Watch Tutorial

Checkout more Solutions here