leetcode

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docs/LeetCode/4sum.md

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+[LeetCode - The World's Leading Online Programming Learning Platform](https://leetcode.com/problems/4sum/)  
+iven an array `nums` of `n` integers, return _an array of all the **unique** quadruplets_ `[nums[a], nums[b], nums[c], nums[d]]` such that:  
+给定一个 `n` 整数数组，返回一个包含所有唯一四胞胎 `[nums[a], nums[b], nums[c], nums[d]]` 的数组 `nums` ，使得：
+
+- `0 <= a, b, c, d < n`
+- `a`, `b`, `c`, and `d` are **distinct**.  
+    `a` 、 `b` `c` 和 `d` 是不同的。
+- `nums[a] + nums[b] + nums[c] + nums[d] == target`
+
+You may return the answer in **any order**.  
+您可以按任何顺序返回答案。
+
+**Example 1: 示例 1：**
+
+**Input:** nums = [1,0,-1,0,-2,2], target = 0
+**Output:** [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
+
+**Example 2: 示例 2：**
+
+**Input:** nums = [2,2,2,2,2], target = 8
+**Output:** [[2,2,2,2]]
+
+**Constraints: 约束：**
+
+- `1 <= nums.length <= 200`
+- `-109 <= nums[i] <= 109`
+- `-109 <= target <= 109`

docs/LeetCode/FizzBuzz.md

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+Given a number n, for each integer
+iin the range from Ito n inclusive,
+print one value per line as follows:
+• If iis a multiple of both 3and 5,
+print FizzBuzz.
+• If iis a multiple of 3(but not 5),
+print Fizz.
+• If iis a multiple of 5(but not 3),
+print BUZZ.
+• If iis nota multiple of 30r 5, print
+the value of i.
+
+Function Description
+Complete the function fizzBuzz in
+he editor below.
+izzBuzz has the following
+parameter(s):
+intn: upper limit of values to test
+inclusive)
+Returns: NONE
+Prints:
+The function must print the
+ppropriate response for each value
+iin the set {1, 2, ... n} in ascending
+order, each on a separate line.
+Constraints
+• 007<2 x 105
+

docs/LeetCode/best-time-to-buy-and-sell-stock-with-transaction-fee.md

@@ -0,0 +1,99 @@
+[LeetCode - The World's Leading Online Programming Learning Platform](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/?envType=study-plan-v2&envId=leetcode-75)  
+
+Companies 公司
+
+You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day, and an integer `fee` representing a transaction fee.  
+您将获得一个数组 `prices` ，其中 `prices[i]` 是给定股票 `ith` 当天的价格，以及一个 `fee` 表示交易费用的整数。
+
+Find the maximum profit you can achieve. You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction.  
+找到您可以实现的最大利润。您可以完成任意数量的交易，但您需要为每笔交易支付交易费用。
+
+**Note: 注意：**
+
+- You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).  
+    您不得同时进行多笔交易（即，您必须在再次买入之前卖出股票）。
+- The transaction fee is only charged once for each stock purchase and sale.  
+    每次股票买卖只收取一次交易费。
+
+**Example 1: 示例 1：**
+
+**Input:** prices = [1,3,2,8,4,9], fee = 2
+**Output:** 8
+**Explanation:** The maximum profit can be achieved by:
+- Buying at prices[0] = 1
+- Selling at prices[3] = 8
+- Buying at prices[4] = 4
+- Selling at prices[5] = 9
+The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
+
+**Example 2: 示例 2：**
+
+**Input:** prices = [1,3,7,5,10,3], fee = 3
+**Output:** 6
+
+**Constraints: 约束：**
+
+- `1 <= prices.length <= 5 * 104`
+- `1 <= prices[i] < 5 * 104`
+- `0 <= fee < 5 * 104`
+
+```java
+package org.leetcode;  
+
+public class MaxProfit {  
+    public int maxProfit(int[] prices, int fee) {  
+        int n = prices.length;  
+        int buy = -prices[0];  
+        // Max money we can have if we have bought a stock on the ith day  
+        int sell = 0;  
+        // Max money we can have if we have sold a stock on the ith day  
+
+        for (int i = 1; i < n; i++) {  
+            int prevBuy = buy;  
+            int prevSell = sell;  
+
+            // Calculate the maximum money we can have if we buy a stock on the current day  
+            buy = Math.max(prevSell - prices[i], prevBuy);  
+            // Calculate the maximum money we can have if we sell a stock on the current day  
+            sell = Math.max(prevBuy + prices[i] - fee, prevSell);  
+        }  
+
+        return sell;  
+
+    }  
+    public static void main(String[] args) {  
+        MaxProfit maxProfit = new MaxProfit();  
+
+        int[] prices1 = {1, 3, 2, 8, 4, 9};  
+        int fee1 = 2;  
+        int result1 = maxProfit.maxProfit(prices1, fee1);  
+        int expect1 = 8;  
+        if (result1 == expect1) {  
+            System.out.println("Pass Max Profit 1: " + result1);  
+        } else {  
+            System.out.println("Fail expected: " + expect1 + " but got: " + result1);  
+        }  
+
+        int[] prices2 = {1, 3, 7, 5, 10, 3};  
+        int fee2 = 3;  
+        int result2 = maxProfit.maxProfit(prices2, fee2);  
+        int expect2 = 6;  
+        if (result2 == expect2) {  
+            System.out.println("Pass Max Profit 2: " + result2);  
+        } else {  
+            System.out.println("Fail expected: " + expect2 + " but got: " + result2);  
+        }  
+
+        int[] prices3 = {4, 6, 2, 8, 3, 10, 14};  
+        int fee3 = 3;  
+        int result3 = maxProfit.maxProfit(prices3, fee3);  
+        int expect3 = 11;  
+        if (result3 == expect3) {  
+            System.out.println("Pass Max Profit 3: " + result3);  
+        } else {  
+            System.out.println("Fail expected: " + expect3 + " but got: " + result3);  
+        }  
+    }  
+
+}
+```

docs/LeetCode/confirmation-rate.md

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+[LeetCode - The World's Leading Online Programming Learning Platform](https://leetcode.com/problems/confirmation-rate/description/)  
+Table: `Confirmations`
+
++----------------+----------+
+| Column Name    | Type     |
++----------------+----------+
+| user_id        | int      |
+| time_stamp     | datetime |
+| action         | ENUM     |
++----------------+----------+
+(user_id, time_stamp) is the primary key (combination of columns with unique values) for this table.
+user_id is a foreign key (reference column) to the Signups table.
+action is an ENUM (category) of the type ('confirmed', 'timeout')
+Each row of this table indicates that the user with ID user_id requested a confirmation message at time_stamp and that confirmation message was either confirmed ('confirmed') or expired without confirming ('timeout').
+
+The **confirmation rate** of a user is the number of `'confirmed'` messages divided by the total number of requested confirmation messages. The confirmation rate of a user that did not request any confirmation messages is `0`. Round the confirmation rate to **two decimal** places.
+
+Write a solution to find the **confirmation rate** of each user.
+
+Return the result table in **any order**.
+
+The result format is in the following example.
+```sql
+# Write your MySQL query statement below
+SELECT 
+S.user_id,
+ROUND(IFNULL(SUM(CASE WHEN c.action = 'confirmed' THEN 1 ELSE 0 END) / COUNT(c.action), 0), 2) AS confirmation_rate
+
+FROM Signups S
+LEFT JOIN Confirmations C
+ON S.user_id = C.user_id
+
+GROUP BY user_id
+```