Buy_And_Sell_Stock_IV.py

You are given an integer array prices where prices[i] is the price of a given stock on the ith day.
Design an algorithm to find the maximum profit. You may complete at most k transactions.
Notice that you may not engage in multiple transactions simultaneously 
(i.e., you must sell the stock before you buy again).

Example 1:
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.

Example 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 <= 109
0 <= prices.length <= 104
0 <= prices[i] <= 1000



# My Solution
# O(nk) Time and O(k) Space
class Solution:
    def maxProfit(self, k: int, prices: List[int]) -> int:
        
        n = len(prices)
        
        if n<=1 or k==0:
            return 0
        
        profit = 0
        
        if k >= n//2:
            for i in range(n-1):
                if prices[i] < prices[i+1]:
                    profit += prices[i+1] - prices[i]
                    
            return profit
        
        buy  = [float('-inf') for i in range(k)]
        sell = [0 for i in range(k)]
        
        for i in range(n):
            for j in range(k):
                
                buy[j]  = max(buy[j], 0-prices[i] if j==0 else sell[j-1]-prices[i])
                sell[j] = max(sell[j], buy[j]+prices[i])
                
        return sell[-1]



# LeetCode's Solution
# O(nk) Time and O(nk) Space
class Solution:
    def maxProfit(self, k: int, prices: List[int]) -> int:
        n = len(prices)

        # solve special cases
        if not prices or k==0:
            return 0

        if 2*k > n:
            res = 0
            for i, j in zip(prices[1:], prices[:-1]):
                res += max(0, i - j)
            return res

        # dp[i][used_k][ishold] = balance
        # ishold: 0 nothold, 1 hold
        dp = [[[-math.inf]*2 for _ in range(k+1)] for _ in range(n)]

        # set starting value
        dp[0][0][0] = 0
        dp[0][1][1] = -prices[0]

        # fill the array
        for i in range(1, n):
            for j in range(k+1):
                # transition equation
                dp[i][j][0] = max(dp[i-1][j][0], dp[i-1][j][1]+prices[i])
                # you can't hold stock without any transaction
                if j > 0:
                    dp[i][j][1] = max(dp[i-1][j][1], dp[i-1][j-1][0]-prices[i])

        res = max(dp[n-1][j][0] for j in range(k+1))
        return res