human
Give me a Python code to find the maximum profit given the following:
Solve in Python:
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 k transactions.

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

Example 1:

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

Example 2:

Input: [3,2,6,5,0,3], k = 2
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.
assistant
#Here is Python code to find the maximum profit.

class Solution:
     def maxProfit(self, k, prices):
         """
         :type k: int
         :type prices: List[int]
         :rtype: int
         """
         '''
         https://www.youtube.com/watch?v=oDhu5uGq_ic
 
             i: number of transactions
             j: jth day 
 
             Total[i][j] = max(T[i][j-1], max((price[j] - price[m]) + T[i-1][m] for m in range(0, j-1)))
             
             NOTEs:
                 1. use cache
                 2. no need to initialize a 2D array, it takes too long
     
         '''
         prices_length = len(prices)
         if prices_length == 0:
             return 0
         if k == 0:
             return 0
 
         # For N days, there will be at most N/2 transactions (N/2 buys and N/2 sells)
         # One buy + One sell = One transaction
         # When k is big enough to allow us buy/sell as many as we want, this becomes type-II 
         # https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/
         if 2*k > (prices_length+1):
             profit = 0
             max_price = prices[0]
             min_price = prices[0]
 
             for p in prices:
                 if p < max_price:
                     profit += (max_price - min_price)
                     max_price = min_price = p
                 else:
                     max_price = p 
                     
             profit += (max_price - min_price)
     
             return profit
 
         # Generic solution 
         candidate_max = float('-inf')
         profit = [0] * prices_length
         for trans in range(1, k+1):
 
             cache = -prices[0]
             max_profit = 0
             for d in range(1, prices_length):
                 if 2*trans > d+1: continue # For a given day, there are too many transaction allowed, no need to compute
 
                 cache = -prices[d-1] + profit[d-1] if -prices[d-1] + profit[d-1] > cache else cache 
                 new_profit = prices[d] + cache 
                 
                 # profit[d-1]...