# Maximum profit by buying and selling a share at most twice

In a daily share trading, a buyer buys shares in the morning and sells it on the same day. If the trader is allowed to make at most 2 transactions in a day, whereas the second transaction can only start after the first one is complete (Sell->buy->sell->buy). Given stock prices throughout the day, find out the maximum profit that a share trader could have made.
Examples:

```Input:   price[] = {10, 22, 5, 75, 65, 80}
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[] = {2, 30, 15, 10, 8, 25, 80}
Output:  100
Trader earns 100 as sum of 28 and 72
Buy at price 2, sell at 30, buy at 8 and sell at 80

Input:   price[] = {100, 30, 15, 10, 8, 25, 80};
Output:  72
Buy at price 8 and sell at 80.

Input:   price[] = {90, 80, 70, 60, 50}
Output:  0
Not possible to earn.

```

A Simple Solution is to consider every index ‘i’ and do following

```Max profit with at most two transactions =
MAX {max profit with one transaction and subarray price[0..i] +
max profit with one transaction and subarray price[i+1..n-1]  }
i varies from 0 to n-1.

```

Maximum possible using one transaction can be calculated using the following O(n) algorithm
The maximum difference between two elements such that larger element appears after the smaller number
Time complexity of above simple solution is O(n2).

We can do this O(n) using following Efficient Solution. The idea is to store the maximum possible profit of every subarray and solve the problem in the following two phases.

1) Create a table profit[0..n-1] and initialize all values in it 0.
2) Traverse price[] from right to left and update profit[i] such that profit[i] stores maximum profit achievable from one transaction in subarray price[i..n-1]
3) Traverse price[] from left to right and update profit[i] such that profit[i] stores maximum profit such that profit[i] contains maximum achievable profit from two transactions in subarray price[0..i].
4) Return profit[n-1]

To do step 2, we need to keep track of the maximum price from right to left side and to do step 3, we need to keep track of the minimum price from left to right. Why we traverse in reverse directions? The idea is to save space, in the third step, we use the same array for both purposes, maximum with 1 transaction and maximum with 2 transactions. After iteration i, the array profit[0..i] contains the maximum profit with 2 transactions and profit[i+1..n-1] contains profit with two transactions.

Below are the implementations of the above idea.

## C++

 `// C++ program to find maximum possible profit with at most` `// two transactions` `#include` `using` `namespace` `std;`   `// Returns maximum profit with two transactions on a given` `// list of stock prices, price[0..n-1]` `int` `maxProfit(``int` `price[], ``int` `n)` `{` `    ``// Create profit array and initialize it as 0` `    ``int` `*profit = ``new` `int``[n];` `    ``for` `(``int` `i=0; i=0;i--)` `    ``{` `        ``// max_price has maximum of price[i..n-1]` `        ``if` `(price[i] > max_price)` `            ``max_price = price[i];`   `        ``// we can get profit[i] by taking maximum of:` `        ``// a) previous maximum, i.e., profit[i+1]` `        ``// b) profit by buying at price[i] and selling at` `        ``//    max_price` `        ``profit[i] = max(profit[i+1], max_price-price[i]);` `    ``}`   `    ``/* Get the maximum profit with two transactions allowed` `       ``After this loop, profit[n-1] contains the result */` `    ``int` `min_price = price;` `    ``for` `(``int` `i=1; i

## Java

 `class` `Profit` `{` `    ``// Returns maximum profit with two transactions on a given` `    ``// list of stock prices, price[0..n-1]` `    ``static` `int` `maxProfit(``int` `price[], ``int` `n)` `    ``{` `        ``// Create profit array and initialize it as 0` `        ``int` `profit[] = ``new` `int``[n];` `        ``for` `(``int` `i=``0``; i=``0``;i--)` `        ``{` `            ``// max_price has maximum of price[i..n-1]` `            ``if` `(price[i] > max_price)` `                ``max_price = price[i];` `     `  `            ``// we can get profit[i] by taking maximum of:` `            ``// a) previous maximum, i.e., profit[i+1]` `            ``// b) profit by buying at price[i] and selling at` `            ``//    max_price` `            ``profit[i] = Math.max(profit[i+``1``], max_price-price[i]);` `        ``}` `     `  `        ``/* Get the maximum profit with two transactions allowed` `           ``After this loop, profit[n-1] contains the result */` `        ``int` `min_price = price[``0``];` `        ``for` `(``int` `i=``1``; i

## Python

 `# Returns maximum profit with two transactions on a given ` `# list of stock prices price[0..n-1]` `def` `maxProfit(price,n):` `    `  `    ``# Create profit array and initialize it as 0` `    ``profit ``=` `[``0``]``*``n` `    `  `    ``# Get the maximum profit with only one transaction` `    ``# allowed. After this loop, profit[i] contains maximum` `    ``# profit from price[i..n-1] using at most one trans.` `    ``max_price``=``price[n``-``1``]` `    `  `    ``for` `i ``in` `range``( n``-``2``, ``0` `,``-``1``):` `        `  `        ``if` `price[i]> max_price:` `            ``max_price ``=` `price[i]` `            `  `        ``# we can get profit[i] by taking maximum of:` `        ``# a) previous maximum, i.e., profit[i+1]` `        ``# b) profit by buying at price[i] and selling at` `        ``#    max_price` `        ``profit[i] ``=` `max``(profit[i``+``1``], max_price ``-` `price[i])` `        `  `    ``# Get the maximum profit with two transactions allowed` `    ``# After this loop, profit[n-1] contains the result    ` `    ``min_price``=``price[``0``]` `    `  `    ``for` `i ``in` `range``(``1``,n):` `        `  `        ``if` `price[i] < min_price:` `            ``min_price ``=` `price[i]`   `        ``# Maximum profit is maximum of:` `        ``# a) previous maximum, i.e., profit[i-1]` `        ``# b) (Buy, Sell) at (min_price, A[i]) and add` `        ``#    profit of other trans. stored in profit[i]    ` `        ``profit[i] ``=` `max``(profit[i``-``1``], profit[i]``+``(price[i]``-``min_price))` `        `  `    ``result ``=` `profit[n``-``1``]` `    `  `    ``return` `result`   `# Driver function` `price ``=` `[``2``, ``30``, ``15``, ``10``, ``8``, ``25``, ``80``]` `print` `"Maximum profit is"``, maxProfit(price, ``len``(price))`   `# This code is contributed by __Devesh Agrawal__`

## C#

 `// C# program to find maximum possible profit` `// with at most two transactions` `using` `System;`   `class` `GFG {` `    `  `    ``// Returns maximum profit with two` `    ``// transactions on a given list of ` `    ``// stock prices, price[0..n-1]` `    ``static` `int` `maxProfit(``int` `[]price, ``int` `n)` `    ``{` `        `  `        ``// Create profit array and initialize` `        ``// it as 0` `        ``int` `[]profit = ``new` `int``[n];` `        ``for` `(``int` `i = 0; i < n; i++)` `            ``profit[i] = 0;` `    `  `        ``/* Get the maximum profit with only` `        ``one transaction allowed. After this ` `        ``loop, profit[i] contains maximum` `        ``profit from price[i..n-1] using at` `        ``most one trans. */` `        ``int` `max_price = price[n-1];` `        `  `        ``for` `(``int` `i = n-2; i >= 0; i--)` `        ``{` `            `  `            ``// max_price has maximum of` `            ``// price[i..n-1]` `            ``if` `(price[i] > max_price)` `                ``max_price = price[i];` `    `  `            ``// we can get profit[i] by taking ` `            ``// maximum of:` `            ``// a) previous maximum, i.e., ` `            ``// profit[i+1]` `            ``// b) profit by buying at price[i]` `            ``// and selling at max_price` `            ``profit[i] = Math.Max(profit[i+1], ` `                          ``max_price - price[i]);` `        ``}` `    `  `        ``/* Get the maximum profit with two` `        ``transactions allowed After this loop,` `        ``profit[n-1] contains the result */` `        ``int` `min_price = price;` `        `  `        ``for` `(``int` `i = 1; i < n; i++)` `        ``{` `            `  `            ``// min_price is minimum price in` `            ``// price[0..i]` `            ``if` `(price[i] < min_price)` `                ``min_price = price[i];` `    `  `            ``// Maximum profit is maximum of:` `            ``// a) previous maximum, i.e.,` `            ``// profit[i-1]` `            ``// b) (Buy, Sell) at (min_price,` `            ``// price[i]) and add profit of ` `            ``// other trans. stored in` `            ``// profit[i]` `            ``profit[i] = Math.Max(profit[i-1],` `                       ``profit[i] + (price[i]` `                              ``- min_price) );` `        ``}` `        ``int` `result = profit[n-1];` `        `  `        ``return` `result;` `    ``}`     `    ``public` `static` `void` `Main()` `    ``{` `        ``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 nitin mittal.`

## PHP

 `= 0; ``\$i``--) ` `    ``{ ` `        ``// max_price has maximum ` `        ``// of price[i..n-1] ` `        ``if` `(``\$price``[``\$i``] > ``\$max_price``) ` `            ``\$max_price` `= ``\$price``[``\$i``]; `   `        ``// we can get profit[i] by ` `        ``// taking maximum of: ` `        ``// a) previous maximum, ` `        ``//    i.e., profit[i+1] ` `        ``// b) profit by buying at ` `        ``// price[i] and selling at ` `        ``// max_price ` `        ``if``(``\$profit``[``\$i` `+ 1] >` `           ``\$max_price``-``\$price``[``\$i``])` `        ``\$profit``[``\$i``] = ``\$profit``[``\$i` `+ 1]; ` `        ``else` `        ``\$profit``[``\$i``] = ``\$max_price` `-` `                      ``\$price``[``\$i``];` `    ``} `   `    ``// Get the maximum profit with ` `    ``// two transactions allowed. ` `    ``// After this loop, profit[n-1] ` `    ``// contains the result ` `    ``\$min_price` `= ``\$price``; ` `    ``for` `(``\$i` `= 1; ``\$i` `< ``\$n``; ``\$i``++) ` `    ``{ ` `        ``// min_price is minimum ` `        ``// price in price[0..i] ` `        ``if` `(``\$price``[``\$i``] < ``\$min_price``) ` `            ``\$min_price` `= ``\$price``[``\$i``]; `   `        ``// Maximum profit is maximum of: ` `        ``// a) previous maximum, ` `        ``//    i.e., profit[i-1] ` `        ``// b) (Buy, Sell) at (min_price, ` `        ``//     price[i]) and add ` `        ``// profit of other trans. ` `        ``// stored in profit[i] ` `        ``\$profit``[``\$i``] = max(``\$profit``[``\$i` `- 1], ` `                          ``\$profit``[``\$i``] + ` `                         ``(``\$price``[``\$i``] - ``\$min_price``)); ` `    ``} ` `    ``\$result` `= ``\$profit``[``\$n` `- 1]; ` `    ``return` `\$result``; ` `}`   `// Driver Code ` `\$price` `= ``array``(2, 30, 15, 10,` `               ``8, 25, 80); ` `\$n` `= sizeof(``\$price``); ` `echo` `"Maximum Profit = "``. ` `      ``maxProfit(``\$price``, ``\$n``); ` `    `  `// This code is contributed` `// by Arnab Kundu` `?>`

Output:

```Maximum Profit = 100

```

The time complexity of the above solution is O(n).