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

Given an array prices[] which denotes the prices of the stocks on different days, the task is to find the maximum profit possible for 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:
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:
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 
    1. Find the minimum price we have to pay from our packet (buy1) to buy 1st stock which is the price of the stock on that day or on any previous day.
    2. Find the maximum profit by selling the 1st stock which on the current day.
  • Maximum profit for second Transaction 
    1. The minimum price we have to pay from our pocket to buy 2 stocks will be 
       
      buy2 = price[i] - profit1, 
      // Profit 1 is the profit from 
      // selling the first stock
      
      
    2. 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 = 2
Now we will buy the 2nd stock on 3rd day
Stock 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) = 2
buy2 = 2
Now, we sell 2nd 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++

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// 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 prices
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_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 Code
int 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;
}

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Java

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// Java implmenetation to find
// maximum possible profit
// with at most two transactions
import java.util.*;
  
class GFG{
  
// Function to find the maximum
// profit with two transactions
// on a given list of stock prices
static 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 Code
public 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

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C#

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// C# implmenetation to find
// maximum possible profit
// with at most two transactions
using System;
  
class GFG{
  
// Function to find the maximum
// profit with two transactions
// on a given list of stock prices
static 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 Code
public 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

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Output: 

Maximum Profit = 100

 

Time complexity: O(N) 
Auxiliary Space: O(1)
 

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