## Algorithm

Problem Name: 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
```

Example 2:

```Input: prices = 
Output: 0
```

Constraints:

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

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

``````
int _max(int a, int b) {
return a > b ? a : b;
}
int maxProfit(int* prices, int pricesSize) {
int i;
int buy_today, sell_today, cool_today;  // max profit of every possible status on a day

sell_today = cool_today = 0;

for (i = 1; i < pricesSize; i ++) {
sell_yesterday = sell_today;
cool_yesterday = cool_today;

sell_today = _max(sell_yesterday, buy_yesterday + prices[i]);
cool_today = _max(cool_yesterday, sell_yesterday);
}

return sell_today;
}
``````
Copy The Code &

Input

cmd
prices = [1,2,3,0,2]

Output

cmd
3

### #2 Code Example with C++ Programming

```Code - C++ Programming```

``````
class Solution {
public:
int maxProfit(vector& prices) {
int n = prices.size();
if(n < 2) return 0;
sell = 0;
rest = 0;
for(int i = 1; i < n; i++){
sell[i] = max(sell[i - 1], buy[i - 1] + prices[i]);
rest[i] = max(rest[i - 1], sell[i - 1]);
}
return max(rest[n - 1], sell[n - 1]);
}
};
``````
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Input

cmd
prices = [1,2,3,0,2]

Output

cmd
3

### #3 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
int[] prices;
Integer[] sellCache;
public int maxProfit(int[] prices) {
this.prices = prices;
this.sellCache = new Integer[prices.length];
}

if (idx >= prices.length) {
return 0;
}
}
int cost = -prices[idx];
int bestProfitBuying = sell(idx + 1) + cost;
}

private int sell(int idx) {
if (idx == prices.length) {
return 0;
}
if (sellCache[idx] != null) {
return sellCache[idx];
}
int price = prices[idx];
int bestProftSelling = buy(idx + 2) + price;
int bestProfitNotSelling = sell(idx + 1);
sellCache[idx] = Math.max(bestProftSelling, bestProfitNotSelling);
return sellCache[idx];
}
}
``````
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Input

cmd
prices = 

### #4 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const maxProfit = function(prices) {
if (prices === null || prices.length < 1) {
return 0;
}

const length = prices.length;
// buy[i]: max profit if the first "i" days end with a "buy" day
const buy = Array(length + 1).fill(0);
// buy[i]: max profit if the first "i" days end with a "sell" day
const sell = Array(length + 1).fill(0);

for (let i = 2; i <= length; i++) {
const price = prices[i - 1];
sell[i] = Math.max(sell[i - 1], buy[i - 1] + price);
}

return sell[length];
};
``````
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Input

cmd
prices = 

### #5 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def maxProfit(self, prices):
dp1, dp2, dp3 = 0, 0, -float("inf")
for p in prices:
dp1, dp2, dp3 = dp3 + p, max(dp1, dp2), max(dp2 - p, dp3)
return max(dp1, dp2)
``````
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Input

cmd
prices = [1,2,3,0,2]

Output

cmd
3

### #6 Code Example with C# Programming

```Code - C# Programming```

``````
using System;

namespace LeetCode
{
{
public int MaxProfit(int[] prices)
{
int sold = int.MinValue, held = int.MinValue, reset = 0;

foreach (var price in prices)
{
var temp = sold;

sold = held + price;
held = Math.Max(held, reset - price);
reset = Math.Max(reset, temp);
}

return Math.Max(sold, reset);
}
}
}
``````
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Input

cmd
prices = [1,2,3,0,2]

Output

cmd
3