Most_Imp_Interview_leetcode_Questions_150/DP(MultiDimenisonal)/BestTimeToBuyAndSellStockIII.py at main · Ak-Rajak/Most_Imp_Interview_leetcode_Questions_150 · GitHub

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# Question link - https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/description/?envType=study-plan-v2&envId=top-interview-150

class Solution:
    # this the solution for the space optimization problem
    # Complexity T(N*2*3) SC: O(2*3)
    def maxProfit(self, prices: List[int]) -> int:
        n = len(prices)
        if n == 0:
            return 0

        # # 3D Dp for memoziarion
        # dp = [[[0 for _ in range(3)] for _ in range(2)] for _ in range(n + 1)]

        # Conver the 3D into 2D
        cur = [[0] * 3 for _ in range(2)]
        ahead = [[0] * 3 for _ in range(2)]


        # Traverse the dp in 3D space
        for ind in range(n - 1, -1, -1):
            for buy in range(2):
                # Cap == 0 is not need to compute
                for cap in range(1,3):

                    # Recurence relations
                    if buy == 0:
                        cur[buy][cap] = max(-prices[ind] + ahead[1][cap] , ahead[0][cap])
                    if buy == 1:
                        cur[buy][cap] = max(prices[ind] + ahead[0][cap-1] ,ahead[1][cap])

            # Upadte the ahead with cur
            ahead = [row[:] for row in cur]

        # Return the result
        return ahead[0][2]


    # this is the tabulation problem
    # The solution for the problem with complexity T(N*2*3)
    # Space Complexity : O(N*2*3)
    # def maxProfit(self, prices: List[int]) -> int:
    #     n = len(prices)
    #     if n == 0:
    #         return 0

    #     # 3D Dp for memoziarion
    #     dp = [[[0 for _ in range(3)] for _ in range(2)] for _ in range(n + 1)]

    #     # Traverse the dp in 3D space
    #     for ind in range(n - 1, -1, -1):
    #         for buy in range(2):
    #             # Cap == 0 is not need to compute
    #             for cap in range(1,3):

    #                 # Recurence relations
    #                 if buy == 0:
    #                     dp[ind][buy][cap] = max(-prices[ind] + dp[ind + 1][1][cap] , dp[ind+1][0][cap])
    #                 if buy == 1:
    #                     dp[ind][buy][cap] = max(prices[ind] + dp[ind + 1][0][cap-1] ,dp[ind+1][1][cap])

    #     # Return the result
    #     return dp[0][0][2]


    # this is the DP memorization approch With complexity : T(N*)
    # TC : O(N*2*3) SC: O(N*2*3) + O(N) for recursion stack
    # def maxProfit(self, prices: List[int]) -> int:
    #     n = len(prices)
    #     if n == 0:
    #         return 0

    #     # 3D Dp for memoziarion
    #     dp = [[[-1 for _ in range(3)] for _ in range(2)] for _ in range(n)]


    #     # Recusive function
    #     def maxProfitOfBTBS(ind,buy,cap,n, prices,dp):
    #         # Base case
    #         if ind == n or cap == 0:
    #             return 0

    #         # If the value is  cahced inside the DP
    #         if dp[ind][buy][cap] != -1:
    #             return dp[ind][buy][cap]

    #         profit = 0

    #         # Case for the buy
    #         if buy == 0:
    #             profit = max(-prices[ind] + maxProfitOfBTBS(ind + 1 ,1,cap,n,prices,dp) , 0 + maxProfitOfBTBS(ind + 1,0,cap,n,prices,dp))
    #         else:
    #             profit = max(prices[ind] + maxProfitOfBTBS(ind + 1,0,cap-1,n,prices,dp), 0 + maxProfitOfBTBS(ind + 1,1,cap,n,prices,dp))
    #         return profit
    #     #Call the recursive function
    #     ans = maxProfitOfBTBS(0,0,2,n,prices,dp)
    #     return ans


    # def maxProfit(self, prices: List[int]) -> int:
    #     n = len(prices)
    #     if n == 0:
    #         return 0

    #     # Complexity TC: O(2^n) SC: O(n)
    #     # Recusive function
    #     def maxProfitOfBTBS(ind,buy,cap,n, prices):
    #         # Base case
    #         if ind == n or cap == 0:
    #             return 0

    #         profit = 0

    #         # Case for the buy
    #         if buy == 0:
    #             profit = max(-prices[ind] + maxProfitOfBTBS(ind + 1 ,1,cap,n,prices) , 0 + maxProfitOfBTBS(ind + 1,0,cap,n,prices))
    #         else:
    #             profit = max(prices[ind] + maxProfitOfBTBS(ind + 1,0,cap-1,n,prices), 0 + maxProfitOfBTBS(ind + 1,1,cap,n,prices))

    #         return profit

    #     #Call the recursive function
    #     ans = maxProfitOfBTBS(0,0,2,n,prices)
    #     return ans