123. Best Time to Buy and Sell Stock III LeetCode Solution

In this guide, you will get 123. Best Time to Buy and Sell Stock III LeetCode Solution with the best time and space complexity. The solution to Best Time to Buy and Sell Stock III problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Best Time to Buy and Sell Stock III solution in C++
Best Time to Buy and Sell Stock III solution in Java
Best Time to Buy and Sell Stock III solution in Python
Additional Resources

123. Best Time to Buy and Sell Stock III LeetCode Solution image

Problem Statement of Best Time to Buy and Sell Stock III

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.

Constraints:

1 <= prices.length <= 105
0 <= prices[i] <= 105

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(1)

123. Best Time to Buy and Sell Stock III LeetCode Solution in C++

class Solution {
 public:
  int maxProfit(vector<int>& 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;
  }
};
/* code provided by PROGIEZ */

123. Best Time to Buy and Sell Stock III LeetCode 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;
  }
}
// code provided by PROGIEZ

123. Best Time to Buy and Sell Stock III LeetCode Solution in Python

class Solution:
  def maxProfit(self, prices: list[int]) -> int:
    sellTwo = 0
    holdTwo = -math.inf
    sellOne = 0
    holdOne = -math.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
 # code by PROGIEZ