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

08 Jul, 2022
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 – 
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.
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 daysint 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 codeint 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;}
// Java program to find out maximum// profit by buying and selling a// share atmost k times given// stock price of n days class GFG {         // Function to find out    // maximum profit by    // buying & selling a    // share atmost k times    // given stock price of n days    static 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 = new int[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 = 0;                 for (int m = 0; m < j; m++)                max_so_far = Math.max(max_so_far, price[j] -                             price[m] + profit[i - 1][m]);                 profit[i][j] = Math.max(profit[i] [j - 1],                                              max_so_far);            }        }         return profit[k][n - 1];    }     // Driver code    public static void main(String []args)    {        int k = 2;        int[] price = { 10, 22, 5, 75, 65, 80 };        int n = price.length;        System.out.println("Maximum profit is: " +                            maxProfit(price, n, k));    }} // This code is contributed by Anshul Aggarwal.
# Python program to maximize the profit# by doing at most k transactions# given stock prices for n days # Function to find out maximum profit by# buying & selling a share atmost k times# given stock price of n daysdef maxProfit(prices, n, k):         # Bottom-up DP approach    profit = [[0 for i in range(k + 1)]                 for j in range(n)]         # Profit is zero for the first    # day and for zero transactions    for i in range(1, n):                 for j in range(1, k + 1):            max_so_far = 0                         for l in range(i):                max_so_far = max(max_so_far, prices[i] -                            prices[l] + profit[l][j - 1])                                         profit[i][j] = max(profit[i - 1][j], max_so_far)         return profit[n - 1][k] # Driver codek = 2prices = [10, 22, 5, 75, 65, 80]n = len(prices) print("Maximum profit is:",       maxProfit(prices, n, k)) # This code is contributed by vaibhav29498
// C# program to find out maximum profit by// buying and selling a share atmost k times// given stock price of n daysusing System; class GFG {         // Function to find out maximum profit by    // buying & selling/ a share atmost k times    // given stock price of n days    static 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 = new int[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 = 0;                 for (int m = 0; m < j; m++)                max_so_far = Math.Max(max_so_far, price[j] -                               price[m] + profit[i - 1, m]);                 profit[i, j] = Math.Max(profit[i, j - 1], max_so_far);            }        }         return profit[k, n - 1];    }     // Driver code to test above    public static void Main()    {        int k = 2;        int[] price = { 10, 22, 5, 75, 65, 80 };         int n = price.Length;         Console.Write("Maximum profit is: " +                     maxProfit(price, n, k));    }} // This code is contributed by Sam007
<?php// PHP program to find out maximum profit by// buying and selling a share atmost k times// given stock price of n days // Function to find out maximum profit by buying// & selling a share atmost k times given stock// price of n daysfunction maxProfit($price, $n, $k){         // table to store results of subproblems    // profit[t][i] stores maximum profit using    // atmost t transactions up to day i (including    // day i)    $profit[$k + 1][$n + 1] = 0;         // For day 0, you can't earn money    // irrespective of how many times you trade    for ($i = 0; $i <= $k; $i++)        $profit[$i][0] = 0;     // profit is 0 if we don't    // do any transaction    // (i.e. k =0)    for ($j = 0; $j <= $n; $j++)        $profit[0][$j] = 0;     // fill the table in    // bottom-up fashion    for ($i = 1; $i <= $k; $i++) {        for ($j = 1; $j < $n; $j++) {            $max_so_far = PHP_INT_MIN;             for ($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    $k = 2;    $price = array (10, 22, 5, 75, 65, 80 );    $n = sizeof($price);    echo "Maximum profit is: ". maxProfit($price, $n, $k); // This is contributed by mits?>
<script> // javascript program to find out maximum// profit by buying and selling a// share atmost k times given// stock price of n days  // Function to find out// maximum profit by// buying & selling a// share atmost k times// given stock price of n daysfunction maxProfit(price,n, k){         // table to store results    // of subproblems    // profit[t][i] stores    // maximum profit using    // atmost t transactions up    // to day i (including day i)    var profit = Array(k+1).fill(0).map    (x => Array(n+1).fill(0));     // For day 0, you can't    // earn money irrespective    // of how many times you trade    for (i = 0; i <= k; i++)        profit[i][0] = 0;     // profit is 0 if we don't    // do any transaction    // (i.e. k =0)    for (j = 0; j <= n; j++)        profit[0][j] = 0;     // fill the table in    // bottom-up fashion    for (i = 1; i <= k; i++)    {        for (j = 1; j < n; j++)        {            var max_so_far = 0;             for (m = 0; m < j; m++)            max_so_far = Math.max(max_so_far, price[j] -                         price[m] + profit[i - 1][m]);             profit[i][j] = Math.max(profit[i] [j - 1],                                          max_so_far);        }    }     return profit[k][n - 1];} // Driver codevar k = 2;var price = [ 10, 22, 5, 75, 65, 80 ];var n = price.length;document.write("Maximum profit is: " +                    maxProfit(price, n, k));  // This code contributed by shikhasingrajput </script>
Output : 
Maximum profit is: 87
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 daysint 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 Codeint 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;}
// Java program to find out maximum// profit by buying and selling a// share atmost k times given stock// price of n daysimport java.io.*; class GFG {         // Function to find out maximum profit by    // buying & selling/ a share atmost k times    // given stock price of n days    static 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[][] = new int[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 = Integer.MIN_VALUE;            for (int j = 1; j < n; j++)            {                prevDiff = Math.max(prevDiff,                           profit[i - 1][j - 1] -                           price[j - 1]);                profit[i][j] = Math.max(profit[i][j - 1],                               price[j] + prevDiff);            }        }         return profit[k][n - 1];    } // Driver codepublic static void main (String[] args)    {        int k = 3;        int price[] = {12, 14, 17, 10, 14, 13, 12, 15};         int n = price.length;         System.out.println("Maximum profit is: " +                            maxProfit(price, n, k));    }} // This code is contributed by Sam007
# Python3 program to find out maximum# profit by buying and/ selling a share# at most k times given the stock price# of n days # Function to find out maximum profit# by buying & selling/ a share atmost# k times given stock price of n daysdef maxProfit(price, n, k):     # Table to store results of subproblems    # profit[t][i] stores maximum profit    # using atmost t transactions up to    # day i (including day i)    profit = [[0 for i in range(n + 1)]                 for j in range(k + 1)]     # Fill the table in bottom-up fashion    for i in range(1, k + 1):        prevDiff = float('-inf')                 for j in range(1, n):            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 Codeif __name__ == "__main__":     k = 3    price = [12, 14, 17, 10, 14, 13, 12, 15]    n = len(price)     print("Maximum profit is:",           maxProfit(price, n, k)) # This code is contributed# by Rituraj Jain
// C# program to find out maximum profit by buying// and selling a share atmost k times given stock// price of n daysusing System; class GFG {         // Function to find out maximum profit by    // buying & selling/ a share atmost k times    // given stock price of n days    static 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 = new int[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.MinValue;            for (int j = 1; j < n; j++) {            prevDiff = Math.Max(prevDiff,                            profit[i - 1, j - 1] - price[j - 1]);            profit[i, j] = Math.Max(profit[i, j - 1],                                price[j] + prevDiff);            }        }         return profit[k, n - 1];    }     // Driver code to test above    public static void Main()    {        int k = 3;        int[] price = {12, 14, 17, 10, 14, 13, 12, 15};         int n = price.Length;         Console.Write("Maximum profit is: " +                     maxProfit(price, n, k));    }} // This code is contributed by Sam007
<?php// PHP program to find out maximum// profit by buying and selling a// share atmost k times given stock// price of n days // Function to find out maximum// profit by buying & selling a// share atmost k times given// stock price of n daysfunction maxProfit($price, $n, $k){         // table to store results    // of subproblems profit[t][i]    // stores maximum profit using    // atmost t transactions up to    // day i (including day i)    $profit[$k + 1][$n + 1]=0;     // For day 0, you can't    // earn money irrespective    // of how many times you trade    for ($i = 0; $i <= $k; $i++)        $profit[$i][0] = 0;     // profit is 0 if we don't    // do any transaction    // (i.e. k =0)    for ($j = 0; $j <= $n; $j++)        $profit[0][$j] = 0;     // fill the table in    // bottom-up fashion    $prevDiff = NULL;    for ($i = 1; $i <= $k; $i++) {        for ($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    $k = 3;    $price = array(12, 14, 17, 10,                   14, 13, 12, 15);    $n = sizeof($price);    echo "Maximum profit is: "         , maxProfit($price, $n, $k); // This code is contributed by nitin mittal?>
<script> // javascript program to find out maximum// profit by buying and selling a// share atmost k times given stock// price of n days        // Function to find out maximum profit by    // buying & selling/ a share atmost k times    // given stock price of n daysfunction maxProfit(price, n , k){         // table to store results of subproblems    // profit[t][i] stores maximum profit    // using atmost t transactions up to day    // i (including day i)    var profit = Array(k+1).fill(0).map(x => Array(n+1).fill(0));     // For day 0, you can't earn money    // irrespective of how many times you trade    for (var i = 0; i <= k; i++)        profit[i][0] = 0;     // profit is 0 if we don't do any    // transaction (i.e. k =0)    for (j = 0; j <= n; j++)        profit[0][j] = 0;     // fill the table in bottom-up fashion    for (var i = 1; i <= k; i++)    {        var prevDiff = -Number.MAX_VALUE;        for (var j = 1; j < n; j++)        {            prevDiff = Math.max(prevDiff,                       profit[i - 1][j - 1] -                       price[j - 1]);            profit[i][j] = Math.max(profit[i][j - 1],                           price[j] + prevDiff);        }    }     return profit[k][n - 1];} // Driver codevar k = 3;var price = [12, 14, 17, 10, 14, 13, 12, 15]; var n = price.length; document.write("Maximum profit is: " +                    maxProfit(price, n, k));// This code contributed by shikhasingrajput</script>
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. 
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:
C++
Javascript
// C++ program to find out maximum profit by// buying and selling a share atmost k times// given stock price of n days#include <bits/stdc++.h>using namespace std; int B;vector<int> A;int dp[501][201][2];int solve(int j, int i, int b){    // if the result has already been calculated return that result    if (dp[i][j][b] != -1)        return dp[i][j][b];    // if i has reached the end of the array return 0    if (i == B)        return 0;    // if we have exhausted the number of transaction return 0    if (j == 0)        return 0;    int res;    // if we are to buy stocks    if (b == 1)        res = max(-A[i] + solve(j, i + 1, 0), solve(j, i + 1, 1));    // if we are to sell stock and complete onr transaction    else        res = max(A[i] + solve(j - 1, i + 1, 1), solve(j, i + 1, 0));    // return the result    return dp[i][j][b] = res;}int maxProfit(int K, int N, int C[]){    A = vector<int>(N, 0);    // Copying C to global A    for (int i = 0; i < N; i++)        A[i] = C[i];    // Initializing DP with -1    for (int i = 0; i <= N; i++)        for (int j = 0; j <= K; j++)        {            dp[i][j][1] = -1;            dp[i][j][0] = -1;        }    // Copying n to global B    B = N;    return solve(K, 0, 1);} // driver codeint main(){    // TEST 1    int k1 = 3;    int price1[] = {12, 14, 17, 10, 14, 13, 12, 15};    int n1 = sizeof(price1) / sizeof(price1[0]);    cout << "Maximum profit is: "         << maxProfit(k1, n1, price1) << endl;    // TEST 2    int k2 = 2;    int price2[] = {10, 22, 5, 75, 65, 80};    int n2 = sizeof(price2) / sizeof(price2[0]);     cout << "Maximum profit is: "         << maxProfit(k2, n2, price2);    return 0;}//This code is contributed by Anirudh Singh
// JavaScript program to find out maximum profit by// buying and selling a share atmost k times// given stock price of n days let B;let A = [];let dp = new Array(501);for (let i = 0; i < 501; i++){    dp[i] = new Array(201);    for (let j = 0; j < 201; j++)        dp[i][j] = new Array(2);} function solve(j, i, b){    // if the result has already been calculated return that result    if (dp[i][j][b] != -1)        return dp[i][j][b];    // if i has reached the end of the array return 0    if (i == B)        return 0;    // if we have exhausted the number of transaction return 0    if (j == 0)        return 0;    let res;    // if we are to buy stocks    if (b == 1)        res = Math.max(-A[i] + solve(j, i + 1, 0), solve(j, i + 1, 1));    // if we are to sell stock and complete onr transaction    else        res = Math.max(A[i] + solve(j - 1, i + 1, 1), solve(j, i + 1, 0));    // return the result    return dp[i][j][b] = res;} function maxProfit(K, N, C){    A = new Array(N).fill(0);    // Copying C to global A    for (i = 0; i < N; i++)        A[i] = C[i];    // Initializing DP with -1    for (i = 0; i <= N; i++)        for (j = 0; j <= K; j++)        {            dp[i][j][1] = -1;            dp[i][j][0] = -1;        }    // Copying n to global B    B = N;    return solve(K, 0, 1);} // driver code // TEST 1let k1 = 3;let price1 = [12, 14, 17, 10, 14, 13, 12, 15];let n1 = price1.length;console.log("Maximum profit is: " + maxProfit(k1, n1, price1));     // TEST 2let k2 = 2;let price2 = [10, 22, 5, 75, 65, 80];let n2 = price2.length; console.log("Maximum profit is: " + maxProfit(k2, n2, price2));          //This code is contributed by phasing17
Maximum profit is: 12
Maximum profit is: 87
Time Complexity: O(n*k)
Auxiliary Space: O(n*k)
 This approach and implementaion is contributed by Anirudh Singh.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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