# 123 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.
``````

Example 4:

``````Input: prices = 
Output: 0
``````

Constraints:

• `1 <= prices.length <= 105`
• `0 <= prices[i] <= 105`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 `````` ``````class Solution: def maxProfit(self, prices: List[int]) -> int: t1_cost, t2_cost = float('inf'), float('inf') t1_profit, t2_profit = 0, 0 for p in prices: # max profit if only one txn is allowed t1_cost = min(t1_cost, p) t1_profit = max(t1_profit, p - t1_cost) # reinvest the gain in the second txn t2_cost = min(t2_cost, p - t1_profit) t2_profit = max(t2_profit, p - t2_cost) return t2_profit``````
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 `````` ``````// dp[k, i] = max(dp[k, i-1], prices[i] - prices[j] + dp[k-1, j-1]), j=[0..i-1] func maxProfit(prices []int) int { if len(prices) == 0 { return 0 } buy1 := prices sell1 := 0 buy2 := prices sell2 := 0 for _, p := range prices { buy1 = min(buy1, p) sell1 = max(sell1, p-buy1) buy2 = min(buy2, p-sell1) sell2 = max(sell2, p-buy2) } return sell2 }``````