add 309

Author: geemaple

README.md

@@ -5,8 +5,8 @@
 My personal leetcode answers<br/>
 This is a **continually updated** open source project<br/>
 <br/>
-Total sovled: **342**<br/>
-Auto updated at: **2024-07-17 17:42:57**<br/>
+Total sovled: **343**<br/>
+Auto updated at: **2024-07-17 19:22:36**<br/>
 
 ## 软件/Softwares
 - [Anki](https://apps.ankiweb.net/)
@@ -195,7 +195,7 @@ python problem.py https://www.lintcode.com/problem/92 -l cpp
 | [Leetcode-3048. Earliest Second To Mark Indices I](https://leetcode.com/problems/earliest-second-to-mark-indices-i/description/) | [python](./leetcode/3048.earliest-second-to-mark-indices-i.py), [c++](./leetcode/3048.earliest-second-to-mark-indices-i.cpp) | O\(M\*logM\) | O\(M\) | \- | - |
 
 ## Dynamic Programming
-| Problem(28) | Solution | Time | Space | Note | Ref |
+| Problem(29) | Solution | Time | Space | Note | Ref |
 | ----- | ----- | ----- | ----- | ----- | ----- |
 | [Leetcode-10. Regular Expression Matching](https://leetcode.com/problems/regular-expression-matching/description/) | [python](./leetcode/10.regular-expression-matching.py), [c++](./leetcode/10.regular-expression-matching.cpp) | O\(MN\) | O\(MN\) | \- | - |
 | [Leetcode-64. Minimum Path Sum](https://leetcode.com/problems/minimum-path-sum/description/) | [c++](./leetcode/64.minimum-path-sum.cpp), [python](./leetcode/64.minimum-path-sum.py) | O\(MN\) | O\(MN\) | \- | - |
@@ -211,6 +211,7 @@ python problem.py https://www.lintcode.com/problem/92 -l cpp
 | [Leetcode-221. Maximal Square](https://leetcode.com/problems/maximal-square/description/) | [python](./leetcode/221.maximal-square.py), [c++](./leetcode/221.maximal-square.cpp) | O\(MN\) | O\(MN\) | \- | - |
 | [Leetcode-279. Perfect Squares](https://leetcode.com/problems/perfect-squares/description/) | [c++](./leetcode/279.perfect-squares.cpp), [python](./leetcode/279.perfect-squares.py) | O\(n^\{3/2\}\) | O\(N\) | \- | - |
 | [Leetcode-300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/description/) | [python](./leetcode/300.longest-increasing-subsequence.py), [c++](./leetcode/300.longest-increasing-subsequence.cpp) | O\(N\*logN\) | O\(N\) | LIS \| std::lower\_bound | - |
+| [Leetcode-309. Best Time To Buy And Sell Stock With Cooldown](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/description/) | [c++](./leetcode/309.best-time-to-buy-and-sell-stock-with-cooldown.cpp), [python](./leetcode/309.best-time-to-buy-and-sell-stock-with-cooldown.py) | O\(N\) | O\(N\) | \- | - |
 | [Leetcode-322. Coin Change](https://leetcode.com/problems/coin-change/description/) | [c++](./leetcode/322.coin-change.cpp), [python](./leetcode/322.coin-change.py) | O\(K \* N\) | O\(N\) | Index | [Video](https://youtu.be/EjMjlFjLRiM) |
 | [Leetcode-375. Guess Number Higher Or Lower II](https://leetcode.com/problems/guess-number-higher-or-lower-ii/description/) | [c++](./leetcode/375.guess-number-higher-or-lower-ii.cpp), [python](./leetcode/375.guess-number-higher-or-lower-ii.py) | O\(N^3\) | O\(N^2\) | DP\(Index\) | - |
 | [Leetcode-413. Arithmetic Slices](https://leetcode.com/problems/arithmetic-slices/description/) | [c++](./leetcode/413.arithmetic-slices.cpp), [python](./leetcode/413.arithmetic-slices.py) | O\(N\) | O\(N\) | \- | - |

leetcode/309.best-time-to-buy-and-sell-stock-with-cooldown.cpp

@@ -0,0 +1,51 @@
+//  Tag: Array, Dynamic Programming
+//  Time: O(N)
+//  Space: O(N)
+//  Ref: -
+//  Note: -
+
+//  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(vector<int>& prices) {
+        int n = prices.size();
+
+        vector<int> rest(n, 0);
+        vector<int> buy(n, 0);
+        vector<int> sell(n, 0);
+
+        buy[0] = -prices[0];
+        for (int i = 1; i < n; i++) {
+            buy[i] = max(buy[i - 1], rest[i - 1] - prices[i]);
+            sell[i] = buy[i] + prices[i];
+            rest[i] = max(rest[i - 1], sell[i - 1]);
+        }
+
+        return max(rest[n - 1], sell[n - 1]);
+    }
+};

leetcode/309.best-time-to-buy-and-sell-stock-with-cooldown.py

@@ -0,0 +1,47 @@
+#  Tag: Array, Dynamic Programming
+#  Time: O(N)
+#  Space: O(N)
+#  Ref: -
+#  Note: -
+
+#  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:
+    def maxProfit(self, prices: List[int]) -> int:
+
+        n = len(prices)
+        rest = [0 for i in range(n)]
+        buy = [0 for i in range(n)]
+        sell = [0 for i in range(n)]
+
+        buy[0] = -prices[0]
+        for i in range(1, n):
+            rest[i] = max(rest[i - 1], sell[i - 1])
+            buy[i] = max(buy[i - 1], rest[i - 1] - prices[i])
+            sell[i] = buy[i] + prices[i]
+            
+        return max(rest[n - 1], sell[n - 1])