Tags: "leetcode", "dynamic-programming", access_time 2-min read

# Best Time to Buy and Sell Stock III

#### Created: November 29, 2019 by [lek-tin]

##### Last updated: November 29, 2019

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most two transactions.

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

### Example 1:

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

### Solution

Dynamic programming

``````class Solution:
def maxProfit(self, prices: List[int]) -> int:
if not prices:
return 0

n = len(prices)
dp = [[0] * n for _ in range(3)]

for i in range(1, 3):
# money spent at first day
prev = dp[i - 1][0] - prices[0]
for j in range(1, n):
# money spent if sell stock today
deal = prev + prices[j]
# compare money spent if don't sell stock today with sell stock today
dp[i][j] = max(dp[i][j - 1], deal)
# compare i - 1 deals during j days, and don't buy stock today
# with i - 1 deals during j days, and buy stock today
prev = max(prev, dp[i - 1][j] - prices[j])

return dp[-1][-1]
``````