LC123.md

Problem: Best Time to Buy and Sell Stock III

Difficulty: Hard

Description:

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).

Examples:

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.

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.

Constraints:

1 <= prices.length <= 10^5
0 <= prices[i] <= 10^5

Solutions:

There are multiple DP approach but I found this approach so simple. We need to have two arrays with the size of k to store minimum price and profits at the ith transaction. So, the first element of the arrays always get upadated per each price in the prices array. This solution has a time complexity of O(Nk) and space complexity of O(k) where N is the size of prices array and k is the number of transactions.

def maxProfit(prices):
    """
    :type prices: List[int]
    :rtype: int
    :Time Complexity: O(N)
    :Space Complexity: O(1)
    """

    if len(prices) == 0:
        return 0
    
    k = 2
    min_price = [float("inf")] * k
    profit = [float("-inf")] * k

    for price in prices:
        min_price[0] = min(min_price[0], price)
        profit[0] = max(profit[0], price - min_price[0])
        for i in range(1, k):
            min_price[i] = min(min_price[i], price - profit[i - 1])
            profit[i] = max(profit[i], price - min_price[i])

    return profit[-1]

For the second solution, we can try to have a 2D array for DP; 1 row per transactions including the transcation zero. Similarly, this solution has a time complexity of O(N) while it has a space complexity of O(N) as well.

def maxProfit(prices):
    """
    :type prices: List[int]
    :rtype: int
    :Time Complexity: O(N)
    :Space Complexity: O(N)
    """
    if len(prices) == 0:
        return 0
    
    k = 2
    max_profit = 0
    dp = [[0 for j in range(len(prices))] for i in range(k + 1)]

    for i in range(1, k + 1):
        maxpr = -prices[0]
        for j in range(1, len(prices)):
            dp[i][j] = max(dp[i][j - 1], maxpr + prices[j])
            maxpr = max(maxpr, -prices[j] + dp[i - 1][j])

    return dp[-1][-1]