309. Best Time to Buy and Sell Stock with Cooldown

1. Description

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 as many transactions as you like (i.e., buy one and sell one share of the stock multiple times) with the following restrictions:

• After you sell your stock, you cannot buy stock on the next day (i.e., cooldown one day).

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

2. Example

Example 1:
Input: prices = [1,2,3,0,2]
Output: 3
Explanation: transactions = [buy, sell, cooldown, buy, sell]

Example 2:
Input: prices = [1]
Output: 0

3. Constraints

• 1 <= prices.length <= 5000
• 0 <= prices[i] <= 1000

4. Solutions

My Accepted Solution

n = prices.size()
Time complexity : O(n)
Space complexity : O(1)

class Solution {
public:
    int maxProfit(const vector<int> &prices) {
        cin.tie(nullptr);
        std::ios::sync_with_stdio(false);

        int max_profits0 = -prices[0]; // 0 means we own a stock
        int max_profits1 = 0;          // 1 means we don't own a stock and in cooldown
        int max_profits2 = 0;          // 2 means we don't own a stock and not in cooldown
        for (int i = 1; i < prices.size(); ++i) {
            tie(max_profits0, max_profits1, max_profits2) = make_tuple(
                max(max_profits0, max_profits2 - prices[i]),
                max_profits0 + prices[i],
                max(max_profits1, max_profits2));
        }

        int result = max(max_profits1, max_profits2);
        return result;
    }
};