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309. Best Time to Buy and Sell Stock with Cooldown

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).

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

Constraints:

  • 1 <= prices.length <= 5000
  • 0 <= prices[i] <= 1000
class Solution {
    public int maxProfit(int[] prices) {
        if (prices.length < 2) {
            return 0;
        }


        int n = prices.length;
        int[] hold = new int[n]; 
        // represents the maximum profit on day i, if we are holding a stock.
        int[] sell = new int[n];
        // represents the maximum profit on day i if we sell a stock
        int[] cooldown = new int[n];
        // represents the maximum profit on day i, if we are in a coolddown period

        hold[0] = -prices[0]; // we bought a stock on day 0
        sell[0] = 0; // we cannot sell on day 0
        cooldown[0] = 0; // we are in cooldown on day 0, no action has been taken

        for (int i = 1; i < n; i++) {
            hold[i] = Math.max(hold[i-1], cooldown[i-1] - prices[i]);
            sell[i] = hold[i-1] + prices[i];
            cooldown[i] = Math.max(cooldown[i-1], sell[i-1]);
        }

        return Math.max(sell[n-1], cooldown[n-1]);
        // the maximum profit will be the maximum value of sell[n-1] and cooldown[n-1]
        // since we cannot hold a stock on the end to consider it as profit.
    }
}

// TC: O(n)
// SC: O(n)

Screenshot 2024-06-28 at 18.31.59