Here, We see Best Time to Buy and Sell Stock with Cooldown LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Best Time to Buy and Sell Stock with Cooldown LeetCode Solution

Problem Statement

You are given an array prices where prices[i] is the price of a given stock on the i^th 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

Best Time to Buy and Sell Stock with Cooldown LeetCode Solution C++

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        int buy(INT_MIN), sell(0), prev_sell(0), prev_buy;
        for (int price : prices) {
            prev_buy = buy;
            buy = max(prev_sell - price, buy);
            prev_sell = sell;
            sell = max(prev_buy + price, sell);
        }
        return sell;
    }
};Code language: PHP (php)

Best Time to Buy and Sell Stock with Cooldown LeetCode Solution Java

class Solution {
    public int maxProfit(int[] prices) {
        int sell = 0, prev_sell = 0, buy = Integer.MIN_VALUE, prev_buy;
        for (int price : prices) {
            prev_buy = buy;
            buy = Math.max(prev_sell - price, prev_buy);
            prev_sell = sell;
            sell = Math.max(prev_buy + price, prev_sell);
        }
        return sell;        
    }
}Code language: JavaScript (javascript)

Best Time to Buy and Sell Stock with Cooldown Solution JavaScript

var maxProfit = function(prices) {
    let [coolDown, sell, hold] = [0, 0, Number.NEGATIVE_INFINITY];
    for( let stockPrice_Day_i of prices){
        let [prevCoolDown, prevSell, prevHold] = [coolDown, sell, hold];
        coolDown = Math.max(prevCoolDown, prevSell);
        sell = prevHold + stockPrice_Day_i;
        hold = Math.max( prevHold, prevCoolDown - stockPrice_Day_i );
    }
    return Math.max(sell, coolDown);
};Code language: JavaScript (javascript)

Best Time to Buy and Sell Stock with Cooldown Solution Python

class Solution(object):
    def maxProfit(self, prices):
        if len(prices) < 2:
            return 0
        sell, buy, prev_sell, prev_buy = 0, -prices[0], 0, 0
        for price in prices:
            prev_buy = buy
            buy = max(prev_sell - price, prev_buy)
            prev_sell = sell
            sell = max(prev_buy + price, prev_sell)
        return sellCode language: HTML, XML (xml)