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

Aditya Goel

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

Difficulty Level : Hard
Last Updated : 28 Jun, 2021

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 = 2Output:  87Trader earns 87 as sum of 12 and 75Buy at price 10, sell at 22, buy at5 and sell at 80Input:Price = [12, 14, 17, 10, 14, 13, 12, 15]K = 3Output:  12Trader earns 12 as the sum of 5, 4 and 3Buy at price 12, sell at 17, buy at 10and sell at 14 and buy at 12 and sellat 15Input:Price = [100, 30, 15, 10, 8, 25, 80]K = 3Output:  72Only one transaction. Buy at price 8and sell at 80.Input:Price = [90, 80, 70, 60, 50]K = 1Output:  0Not 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.

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 –

profit[t][i-1] which represents not doing any transaction on the ith day.
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.

C++
Java
Python3
C#
PHP
Javascript

// C++ program to find out maximum profit by

// buying and selling a share atmost k times

// given stock price of n days

#include <climits>

#include <iostream>

using namespace std;

// Function to find out maximum profit by buying

// & selling a share atmost k times given stock

// price of n days

int maxProfit(int price[], int n, int k)

{

// table to store results of subproblems

// profit[t][i] stores maximum profit using

// atmost t transactions up to day i (including

// day i)

int profit[k + 1][n + 1];

// For day 0, you can't earn money

// irrespective of how many times you trade

for (int i = 0; i <= k; i++)

profit[i][0] = 0;

// profit is 0 if we don't do any transaction

// (i.e. k =0)

for (int j = 0; j <= n; j++)

profit[0][j] = 0;

// fill the table in bottom-up fashion

for (int i = 1; i <= k; i++) {

for (int j = 1; j < n; j++) {

int max_so_far = INT_MIN;

for (int m = 0; m < j; m++)

max_so_far = max(max_so_far,

price[j] - price[m] + profit[i - 1][m]);

profit[i][j] = max(profit[i][j - 1], max_so_far);

}

return profit[k][n - 1];

}

// Driver code

int main()

{

int k = 2;

int price[] = { 10, 22, 5, 75, 65, 80 };

int n = sizeof(price) / sizeof(price[0]);

cout << "Maximum profit is: "

<< maxProfit(price, n, k);

return 0;

}

Output :

Maximum profit is: 87

Optimized Solution:
The above solution has time complexity of 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 –

C++
Java
Python3
C#
PHP
Javascript

// C++ program to find out maximum profit by buying

// and/ selling a share atmost k times given stock

// price of n days

#include <climits>

#include <iostream>

using namespace std;

// Function to find out maximum profit by buying &

// selling/ a share atmost k times given stock price

// of n days

int maxProfit(int price[], int n, int k)

{

// table to store results of subproblems

// profit[t][i] stores maximum profit using atmost

// t transactions up to day i (including day i)

int profit[k + 1][n + 1];

// For day 0, you can't earn money

// irrespective of how many times you trade

for (int i = 0; i <= k; i++)

profit[i][0] = 0;

// profit is 0 if we don't do any transaction

// (i.e. k =0)

for (int j = 0; j <= n; j++)

profit[0][j] = 0;

// fill the table in bottom-up fashion

for (int i = 1; i <= k; i++) {

int prevDiff = INT_MIN;

for (int j = 1; j < n; j++) {

prevDiff = max(prevDiff,

profit[i - 1][j - 1] - price[j - 1]);

profit[i][j] = max(profit[i][j - 1],

price[j] + prevDiff);

}

return profit[k][n - 1];

}

// Driver Code

int main()

{

int k = 3;

int price[] = { 12, 14, 17, 10, 14, 13, 12, 15 };

int n = sizeof(price) / sizeof(price[0]);

cout << "Maximum profit is: "

<< maxProfit(price, n, k);

return 0;

}

Output :

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. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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