Maximum profit by buying and selling a share at most k times

In share trading, a buyer buys shares and sells on a future date. Given the stock price of n days, the trader is allowed to make at most k transactions, where a new transaction can only start after the previous transaction is complete, find out the maximum profit that a share trader could have made. 

Examples:

Input:  
Price = [10, 22, 5, 75, 65, 80]
    K = 2
Output:  87
Trader earns 87 as sum of 12 and 75
Buy at price 10, sell at 22, buy at 
5 and sell at 80
Input:  
Price = [12, 14, 17, 10, 14, 13, 12, 15]
    K = 3
Output:  12
Trader earns 12 as the sum of 5, 4 and 3
Buy at price 12, sell at 17, buy at 10 
and sell at 14 and buy at 12 and sell
at 15
 
Input:  
Price = [100, 30, 15, 10, 8, 25, 80]
    K = 3
Output:  72
Only one transaction. Buy at price 8 
and sell at 80.
Input:  
Price = [90, 80, 70, 60, 50]
    K = 1
Output:  0
Not possible to earn. 

There are various versions of the problem. If we are allowed to buy and sell only once, then we can use the Maximum difference between the two elements algorithm. If we are allowed to make at most 2 transactions, we can follow approach discussed here. If we are allowed to buy and sell any number of times, we can follow approach discussed here.
 

Method 1 Dynamic Programming

In this post, we are only allowed to make at max k transactions. The problem can be solved by using dynamic programming. 

Let profit[t][i] represent maximum profit using at most t transactions up to day i (including day i). Then the relation is:
profit[t][i] = max(profit[t][i-1], max(price[i] – price[j] + profit[t-1][j])) 
          for all j in range [0, i-1] 
profit[t][i] will be maximum of – 

1. profit[t][i-1] which represents not doing any transaction on the ith day.
2. Maximum profit gained by selling on ith day. In order to sell shares on ith day, we need to purchase it on any one of [0, i – 1] days. If we buy shares on jth day and sell it on ith day, max profit will be price[i] – price[j] + profit[t-1][j] where j varies from 0 to i-1. Here profit[t-1][j] is best we could have done with one less transaction till jth day.

Below is Dynamic Programming based implementation

Maximum profit is: 87

Space Complexity: O(n*k) since we are using a 2-D array.
Time Complexity: O(k.n²). It can be reduced if we are able to calculate the maximum profit gained by selling shares on the ith day in constant time.
profit[t][i] = max(profit [t][i-1], max(price[i] – price[j] + profit[t-1][j])) 
                            for all j in range [0, i-1]
If we carefully notice, 
max(price[i] – price[j] + profit[t-1][j]) 
for all j in range [0, i-1]
can be rewritten as, 
= price[i] + max(profit[t-1][j] – price[j]) 
for all j in range [0, i-1] 
= price[i] + max(prevDiff, profit[t-1][i-1] – price[i-1]) 
where prevDiff is max(profit[t-1][j] – price[j]) 
for all j in range [0, i-2]
So, if we have already calculated max(profit[t-1][j] – price[j]) for all j in range [0, i-2], we can calculate it for j = i – 1 in constant time. In other words, we don’t have to look back in the range [0, i-1] anymore to find out best day to buy. We can determine that in constant time using below revised relation.
profit[t][i] = max(profit[t][i-1], price[i] + max(prevDiff, profit [t-1][i-1] – price[i-1]) 
where prevDiff is max(profit[t-1][j] – price[j]) for all j in range [0, i-2]

Below is its optimized implementation – 

Maximum profit is: 12

The time complexity of the above solution is O(kn) and space complexity is O(nk). Space complexity can further be reduced to O(n) as we use the result from the last transaction. But to make the article easily readable, we have used O(kn) space.
This article is contributed by Aditya Goel. 

Method-2 Memoization(dynamic Programming)

One more way we can approach this problem is by considering buying and selling as two different states of the dp. So we will consider the dp as follows:

let i be the index of the stock we are at, j be the total transaction reamaining, b represent if we have to sell or buy this stock(1 for buying and 0 for selling) and A represent the array containing stock prices then state transitions will be as follows:

dp[i][j][1]=max(-A[i]+dp[j][i+1][0],dp[j][i+1][1])

dp[i][j][0]=max(A[i]+dp[j-1][i+1][1],dp[j][i+1][0])

implementation is as follows:

Maximum profit is: 12
Maximum profit is: 87

Time Complexity: O(n*k)
Auxiliary Space: O(n*k), since n*k extra space has been taken.

This approach and implementation is contributed by Anirudh Singh.

Naive Recursive Approach:

Maximum profit is: 87

Time Complexity: O(2 ^(n*k))
Auxiliary Space: O(1), only recursion stack space is used.

Efficient approach : Space Optimized

In this approach we use two vectors: profit and prevDiff. The profit vector stores the maximum profit using at most i transactions up to the current day, and prevDiff vector stores the maximum difference obtained by subtracting the price of the previous day.

By iterating through the prices and transactions in a bottom-up manner, we update the prevDiff and profit vectors accordingly. Finally, we return profit[k], which represents the maximum profit using at most k transactions.

Implementation Steps:

• Define the maxProfit function with parameters price[], n, and k.
• Create vectors profit and prevDiff with initial values.
• Use nested loops to fill the vectors prevDiff and profit in a bottom-up manner.
• Update prevDiff[i] and profit[i] based on the maximum values.
• Return profit[k].

Implementation:

Maximum profit is: 12

Time Complexity: O(n*k)
Auxiliary Space: O(k)

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