LC188.md

Problem: Best Time to Buy and Sell Stock IV

Difficulty: Hard

Description:

You are given an integer array prices where prices[i] is the price of a given stock on the ith day, and an integer k. Find the maximum profit you can achieve. You may complete at most k transactions. Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

Examples:

Input: k = 2, prices = [2,4,1]
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.

Input: k = 2, prices = [3,2,6,5,0,3]
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.

Constraints:

0 <= k <= 100
0 <= prices.length <= 1000
0 <= prices[i] <= 1000

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(k, prices):
    """
    :type k: int
    :type prices: List[int]
    :rtype: int
    :Time Complexity: O(N * k)
    :Space Complexity: O(k)
    """

    if k == 0 or len(prices) == 0:
        return 0

    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(Nk) while it has a space complexity of O(Nk) as well.

def maxProfit(k, prices):
    """
    :type k: int
    :type prices: List[int]
    :rtype: int
    :Time Complexity: O(N * k)
    :Space Complexity: O(N * k)
    """
    if k == 0 or len(prices) == 0:
        return 0

    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]