123. Best Time to Buy and Sell Stock III(Bidirectional Dynamic Programming) 20.10.5 Hard

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: 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 = [1]
Output: 0
 

Constraints:

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


Solution: O(n) T and S
class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        if not prices or len(prices) <= 1:
            return 0
        
        n = len(prices)
        left, right = [0] * n, [0] * (n + 1)
        left_min, right_max = prices[0], prices[-1]
        for i in range(1, n):
            left[i] = max(0, prices[i] - left_min, left[i - 1])
            left_min = min(left_min, prices[i])
        for i in range(n - 2, -1, -1):
            right[i] = max(0, right_max - prices[i], right[i + 1])
            right_max = max(right_max, prices[i])
        max_profit = 0
        for i in range(n):
            max_profit = max(max_profit, left[i] + right[i + 1])
            
        return max_profit