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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Best Time to Buy and Sell Stock III in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemBest Time to Buy and Sell Stock III– LeetCode Problem

Best Time to Buy and Sell Stock III– LeetCode Problem


You are given an array prices where prices[i] is the price of a given stock on the ith day.

Find the maximum profit you can achieve. You may complete at most two transactions.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

Example 1:

Input: prices = [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

Example 2:

Input: prices = [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.


  • 1 <= prices.length <= 105
  • 0 <= prices[i] <= 105
Best Time to Buy and Sell Stock III– LeetCode Solutions
Best Time to Buy and Sell Stock III Solution in C++:
class Solution {
  int maxProfit(vector& prices) {
    int sellTwo = 0;
    int holdTwo = INT_MIN;
    int sellOne = 0;
    int holdOne = INT_MIN;

    for (const int price : prices) {
      sellTwo = max(sellTwo, holdTwo + price);
      holdTwo = max(holdTwo, sellOne - price);
      sellOne = max(sellOne, holdOne + price);
      holdOne = max(holdOne, -price);

    return sellTwo;
Best Time to Buy and Sell Stock III Solution in Java:
class Solution {
  public int maxProfit(int[] prices) {
    int sellTwo = 0;
    int holdTwo = Integer.MIN_VALUE;
    int sellOne = 0;
    int holdOne = Integer.MIN_VALUE;

    for (final int price : prices) {
      sellTwo = Math.max(sellTwo, holdTwo + price);
      holdTwo = Math.max(holdTwo, sellOne - price);
      sellOne = Math.max(sellOne, holdOne + price);
      holdOne = Math.max(holdOne, -price);

    return sellTwo;
Best Time to Buy and Sell Stock III Solution in Python:
class Solution:
  def maxProfit(self, prices: List[int]) -> int:
    sellTwo = 0
    holdTwo = -inf
    sellOne = 0
    holdOne = -inf

    for price in prices:
      sellTwo = max(sellTwo, holdTwo + price)
      holdTwo = max(holdTwo, sellOne - price)
      sellOne = max(sellOne, holdOne + price)
      holdOne = max(holdOne, -price)

    return sellTwo

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