Maximum profit by buying and selling a stock at most twice | Set 2

06 May, 2021
Given an array prices[] which denotes the prices of the stocks on different days, the task is to find the maximum profit possible after buying and selling the stocks on different days using transaction where at most two transactions are allowed.Note: You cannot engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
Examples: 
Input: prices[] = {3, 3, 5, 0, 0, 3, 1, 4} Output: 6 Explanation: Buy on Day 4 and Sell at Day 6 => Profit = 3 – 0 = 3 Buy on Day 7 and Sell at Day 8 => Profit = 4 -1 = 3 Therefore, Total Profit = 3 + 3 = 6
Input: prices[] = {1, 2, 3, 4, 5} Output: 4 Explanation: Buy on Day 1 and sell at Day 6 => Profit = 5 -1 = 4 Therefore, Total Profit = 4 
We have already discussed this problem using Dynamic Programming in O(N) space in this article.Efficient Approach: The idea used here is to keep track of the two transactions at the same time. The two stocks profit is computed as follows 
Maximum Profit for the First Transaction Find the minimum price we have to pay from our pocket (buy1) to get 1st stock which is the price of the stock on that day or on any previous day.Find the maximum profit by selling the 1st stock which on the current day.
Find the minimum price we have to pay from our pocket (buy1) to get 1st stock which is the price of the stock on that day or on any previous day.
Find the maximum profit by selling the 1st stock which on the current day.
Maximum profit for the second Transaction The minimum price we have to pay from our pocket to buy second stocks will be
The minimum price we have to pay from our pocket to buy second stocks will be
buy2 = price[i] - profit1, 
// Profit 1 is the profit from 
// selling the first stock
Find the maximum profit (profit2) by selling the 2nd stock.
Explanation with Example: 
Let us take price[] = { 3, 5, 4, 5} We buy 1st stock on day 1 buy1 = 3.We sell 1st stock on day 2. profit1 = 5 – 3 = 2.We will make a profit of 2 after buying the stock on 1st day and selling on 2nd day. So, profit1 = 2Now we will buy the 2nd stock on 3rd dayStock price on 3rd day is 4 Since we already made a profit of 2, We will spend only 2 extra from our packet to buy the stock i.e., (4 – profit1) = 2buy2 = 2Now, we will sell second stock on day 4.Stock price on 4th day is 5. profit2 = 5 – buy2 = 5 – 2 = 3.Now, we can see that the final profit includes the profit of buying and selling two stocks  
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// C++ implmenetation to find// maximum possible profit// with at most two transactions #include <bits/stdc++.h>using namespace std; // Function to find the maximum// profit with two transactions// on a given list of stock pricesint maxProfit(int price[], int n){     // buy1 - Money lent to buy 1 stock    // profit1 - Profit after selling    // the 1st stock buyed.    // buy2 - Money lent to buy 2 stocks    // including profit of selling 1st stock    // profit2 - Profit after selling 2 stocks    int buy1, profit1, buy2, profit2;     // Set initial buying values to    // INT_MAX as we want to minimize it    buy1 = buy2 = INT_MAX;     // Set initial selling values to    // zero as we want to maximize it    profit1 = profit2 = 0;     for (int i = 0; i < n; i++) {         // Money lent to buy the stock        // should be minimum as possible.        // buy1 tracks the minimum possible        // stock to buy from 0 to i-1.        buy1 = min(buy1, price[i]);         // Profit after selling a stock        // should be maximum as possible.        // profit1 tracks maximum possible        // profit we can make from 0 to i-1.        profit1 = max(profit1, price[i] - buy1);         // Now for buying the 2nd stock,        // we will integrate profit made        // from selling the 1st stock        buy2 = min(buy2, price[i] - profit1);         // Profit after selling a 2 stocks        // should be maximum as possible.        // profit2 tracks maximum possible        // profit we can make from 0 to i-1.        profit2 = max(profit2, price[i] - buy2);    }     return profit2;} // Driver Codeint main(){    int price[] = { 2, 30, 15, 10, 8, 25, 80 };    int n = sizeof(price) / sizeof(price[0]);    cout << "Maximum Profit = "         << maxProfit(price, n);    return 0;}
// Java implmenetation to find// maximum possible profit// with at most two transactionsimport java.util.*; class GFG{ // Function to find the maximum// profit with two transactions// on a given list of stock pricesstatic int maxProfit(int price[], int n){         // buy1 - Money lent to buy 1 stock    // profit1 - Profit after selling    // the 1st stock buyed.    // buy2 - Money lent to buy 2 stocks    // including profit of selling 1st stock    // profit2 - Profit after selling 2 stocks    int buy1, profit1, buy2, profit2;     // Set initial buying values to    // Integer.MAX_VALUE as we want to    // minimize it    buy1 = buy2 = Integer.MAX_VALUE;     // Set initial selling values to    // zero as we want to maximize it    profit1 = profit2 = 0;     for(int i = 0; i < n; i++)    {         // Money lent to buy the stock        // should be minimum as possible.        // buy1 tracks the minimum possible        // stock to buy from 0 to i-1.        buy1 = Math.min(buy1, price[i]);         // Profit after selling a stock        // should be maximum as possible.        // profit1 tracks maximum possible        // profit we can make from 0 to i-1.        profit1 = Math.max(profit1, price[i] - buy1);         // Now for buying the 2nd stock,        // we will integrate profit made        // from selling the 1st stock        buy2 = Math.min(buy2, price[i] - profit1);         // Profit after selling a 2 stocks        // should be maximum as possible.        // profit2 tracks maximum possible        // profit we can make from 0 to i-1.        profit2 = Math.max(profit2, price[i] - buy2);    }    return profit2;} // Driver Codepublic static void main(String[] args){    int price[] = { 2, 30, 15, 10, 8, 25, 80 };    int n = price.length;         System.out.print("Maximum Profit = " +                      maxProfit(price, n));}} // This code is contributed by Rajput-Ji
# Python3 implmenetation to find# maximum possible profit# with at most two transactionsimport sys # Function to find the maximum# profit with two transactions# on a given list of stock pricesdef maxProfit(price, n):     # buy1 - Money lent to buy 1 stock    # profit1 - Profit after selling    # the 1st stock buyed.    # buy2 - Money lent to buy 2 stocks    # including profit of selling 1st stock    # profit2 - Profit after selling 2 stocks         # Set initial buying values to    # INT_MAX as we want to minimize it    buy1, buy2 = sys.maxsize, sys.maxsize       # Set initial selling values to    # zero as we want to maximize it    profit1, profit2 = 0, 0       for i in range(n):           # Money lent to buy the stock        # should be minimum as possible.        # buy1 tracks the minimum possible        # stock to buy from 0 to i-1.        buy1 = min(buy1, price[i])           # Profit after selling a stock        # should be maximum as possible.        # profit1 tracks maximum possible        # profit we can make from 0 to i-1.        profit1 = max(profit1, price[i] - buy1)           # Now for buying the 2nd stock,        # we will integrate profit made        # from selling the 1st stock        buy2 = min(buy2, price[i] - profit1)           # Profit after selling a 2 stocks        # should be maximum as possible.        # profit2 tracks maximum possible        # profit we can make from 0 to i-1.        profit2 = max(profit2, price[i] - buy2)     return profit2   # Driver codeprice = [ 2, 30, 15, 10, 8, 25, 80 ]n = len(price) print("Maximum Profit = ", maxProfit(price, n)) # This code is contributed by divyeshrabadiya07
// C# implmenetation to find// maximum possible profit// with at most two transactionsusing System; class GFG{ // Function to find the maximum// profit with two transactions// on a given list of stock pricesstatic int maxProfit(int []price, int n){         // buy1 - Money lent to buy 1 stock    // profit1 - Profit after selling    // the 1st stock buyed.    // buy2 - Money lent to buy 2 stocks    // including profit of selling 1st stock    // profit2 - Profit after selling 2 stocks    int buy1, profit1, buy2, profit2;     // Set initial buying values to    // int.MaxValue as we want to    // minimize it    buy1 = buy2 = int.MaxValue;     // Set initial selling values to    // zero as we want to maximize it    profit1 = profit2 = 0;     for(int i = 0; i < n; i++)    {         // Money lent to buy the stock        // should be minimum as possible.        // buy1 tracks the minimum possible        // stock to buy from 0 to i-1.        buy1 = Math.Min(buy1, price[i]);         // Profit after selling a stock        // should be maximum as possible.        // profit1 tracks maximum possible        // profit we can make from 0 to i-1.        profit1 = Math.Max(profit1, price[i] - buy1);         // Now for buying the 2nd stock,        // we will integrate profit made        // from selling the 1st stock        buy2 = Math.Min(buy2, price[i] - profit1);         // Profit after selling a 2 stocks        // should be maximum as possible.        // profit2 tracks maximum possible        // profit we can make from 0 to i-1.        profit2 = Math.Max(profit2, price[i] - buy2);    }    return profit2;} // Driver Codepublic static void Main(String[] args){    int []price = { 2, 30, 15, 10, 8, 25, 80 };    int n = price.Length;         Console.Write("Maximum Profit = " +                   maxProfit(price, n));}} // This code is contributed by Rajput-Ji
<script>// javascript implmenetation to find// maximum possible profit// with at most two transactions     // Function to find the maximum    // profit with two transactions    // on a given list of stock prices    function maxProfit(price , n)    {         // buy1 - Money lent to buy 1 stock        // profit1 - Profit after selling        // the 1st stock buyed.        // buy2 - Money lent to buy 2 stocks        // including profit of selling 1st stock        // profit2 - Profit after selling 2 stocks        var buy1, profit1, buy2, profit2;         // Set initial buying values to        // Number.MAX_VALUE as we want to        // minimize it        buy1 = buy2 = Number.MAX_VALUE;         // Set initial selling values to        // zero as we want to maximize it        profit1 = profit2 = 0;         for (i = 0; i < n; i++) {             // Money lent to buy the stock            // should be minimum as possible.            // buy1 tracks the minimum possible            // stock to buy from 0 to i-1.            buy1 = Math.min(buy1, price[i]);             // Profit after selling a stock            // should be maximum as possible.            // profit1 tracks maximum possible            // profit we can make from 0 to i-1.            profit1 = Math.max(profit1, price[i] - buy1);             // Now for buying the 2nd stock,            // we will integrate profit made            // from selling the 1st stock            buy2 = Math.min(buy2, price[i] - profit1);             // Profit after selling a 2 stocks            // should be maximum as possible.            // profit2 tracks maximum possible            // profit we can make from 0 to i-1.            profit2 = Math.max(profit2, price[i] - buy2);        }        return profit2;    }     // Driver Code           var price = [ 2, 30, 15, 10, 8, 25, 80 ];        var n = price.length;         document.write("Maximum Profit = " + maxProfit(price, n)); // This code is contributed by todaysgaurav.</script>
Maximum Profit = 100
 
Time complexity: O(N) Auxiliary Space: O(1)
Rajput-Ji
divyeshrabadiya07
8_bit_spider
todaysgaurav
Arrays
Data Structures
Java 8
Arrays
Dynamic Programming
Data Structures
Arrays
Dynamic Programming
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