Last Updated: 7 Jan, 2021
Best Time to Buy and Sell Stock IV
Hard
Asked in companies
You have been given an array 'PRICES' consisting of 'N' integers where PRICES[i] denotes the price of a given stock on the i-th day. You are also given an integer 'K' denoting the number of possible transactions you can make.
Your task is to find the maximum profit in at most K transactions. A valid transaction involves buying a stock and then selling it.
Note
You can’t engage in multiple transactions simultaneously, i.e. you must sell the stock before rebuying it.
For Example
Input: N = 6 , PRICES = [3, 2, 6, 5, 0, 3] and K = 2.
Output: 7
Explanation : The optimal way to get maximum profit is to buy the stock on day 2(price = 2) and sell it on day 3(price = 6) and rebuy it on day 5(price = 0) and sell it on day 6(price = 3). The maximum profit will be (6 - 2) + (3 - 0) = 7.
Input Format
The first line of input contains an integer 'T' which denotes the number of test cases or queries to be run. Then the test cases follow.
The first line of each test case contains two single-space separated integers ‘N’ and ‘K’, respectively.
The second line of each test case contains ‘N’ single space-separated integers, denoting the elements of the array 'PRICES'.
Output Format
For each test case, print a single integer denoting the maximum profit in at most K transactions.
Print the output of each test case in a separate line.
Note
You do not need to print anything; it has already been taken care of. Just implement the given function.
Constraints
1 <= T <= 100
1 <= N <= 5000
0 <= K <= N/2
0 <= ARR[i] <=10^5
Time Limit : 1 sec
Approaches
The main idea is to use recursion to reduce the big problem into several smaller subproblems.
Algorithm
- We will call a maxProfit function that returns us the maximum profit we can get from the array starting from i (here i denotes the starting index of PRICES) to N - 1. We will use a bool variable buy. If buy == true it denotes that we had bought a stock and now we can’t buy another until we sell it and buy == false denotes that we hadn’t currently bought any stock. The maxProfit function will work as follows:
- If K == 0 || i == N
- Return 0.
- If buy == false
- We make a recursive call to i + 1 with buy marked as false. (Not holding the stock).
- We make a recursive call to i + 1 with buy marked as true and subtract PRICES[i]. (Buy stock on i -th day).
- We finally return the maximum of them.
- Else
- We make a recursive call to i + 1 with buy marked as true. (Hold the stock).
- We make a recursive call to i + 1 with buy marked as false and add PRICES[i] and decrement K by 1. (Sell stock on i-th day).
- Return the maximum of them.
If we draw the recursion tree for the recurrence relation of approach 1, we can observe that there are a lot of overlapping subproblems. There are only N * K distinct recursive calls.
Lets understand this by an example, consider PRICES = [5, 4, 8, 9] and K = 3 and buy == false.
The recursion tree for array will be
As we can clearly see in the recursion tree that [8, 9] for K == 3 and buy == true will be calculated twice.
Since the problem has overlapping subproblems, we can solve it more efficiently using memorization.
The algorithm is similar to the previous approach with some additions.
We initialize a 3-D vector memo with -1. Now, memo[i][j][0] will denote the maximum profit we can get from the array starting from ‘i’ to N - 1 in at most ‘j’ transactions considering currently we haven't bought the stock. memo[i][j][1] will denote the maximum profit we can get from the array starting from ‘i’ to N - 1 in at most ‘j’ transactions considering currently we have bought the stock.
Before calling the function for any valid (i, j, buy), we will check whether we already have a solution corresponding to the (i, j, m) present in memo.
- If we already have a solution corresponding to (i, j, m), we return the solution.
- Else
- If buy == 0
- We call the recursive function for (i + 1, j, 0) and (i + 1, j, 1) - PRICES[i] and store the maximum of them in memo[i][j][buy] and return it.
- Else if buy == 1
- We call the recursive function for (i + 1, j, 1) and (i + 1, j - 1, 0) + PRICES[i] and store the maximum of them in memo[i][j][buy] and return it.
- If buy == 0
Initially, we were breaking the large problem into small subproblems but now, let us look at it differently. The idea is to solve the small problem first and then reach the final answer. Thus we will be using a bottom-up approach now.
We initialize two 2-D matrices, buy[K+1][N+1] and sell[K+1][N+1]. Here, buy[i][j] will denote the maximum profit we can get in at most ‘i’ transactions from the array starting from index 0 and ending at index j -1 and the final state is buying the stock. sell[i][j] will denote the maximum profit we can get in at most ‘i’ transactions from the array starting from index 0 and ending at index j -1 and the final state is not holding the stock i.e. selling the stock.
Algorithm
- We iterate the matrix buy and sell column-wise and update the values.
- buy[j][i] = max(buy[j][i-1], sell[j-1][i-1] - PRICES[i-1]). Here buy[j][j-1] denotes holding the stock and sell[j -1][i - 1] - PRICES[i -1] denotes buying the stock on i-th day.
- sell[j][i] = max(sell[j][i-1], buy[j][i-1]+PRICES[i-1]). Here sell[j][i-1] denotes not holding the stock and buy[i][j - 1] + PRICES[i -1] denotes selling the stock on i-th day.
- Finally return sell[K][N].