Best Time to Buy and Sell Stock III LeetCode Solution

Minimum Cost to Merge Stones
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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

Best Time to Buy and Sell Stock III Solutions

Time: O(n)
Space: O(1)

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;
  }
};

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;
  }
}

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

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