feat: add solutions to lc problems (doocs#1130)

* No.1910.Remove All Occurrences of a Substring
* No.1934.Confirmation Rate
* No.1978.Employees Whose Manager Left the Company
* No.2026.Low-Quality Problems
* No.2175.The Change in Global Rankings
* No.2199.Finding the Topic of Each Post
* No.2230.The Users That Are Eligible for Discount
* No.2480.Form a Chemical Bond

Author: yanglbme

.prettierrc

@@ -32,8 +32,10 @@
                 "solution/1500-1599/1555.Bank Account Summary/Solution.sql",
                 "solution/1600-1699/1667.Fix Names in a Table/Solution.sql",
                 "solution/1700-1799/1777.Product's Price for Each Store/Solution.sql",
+                "solution/1900-1999/1934.Confirmation Rate/Solution.sql",
                 "solution/1900-1999/1972.First and Last Call On the Same Day/Solution.sql",
                 "solution/2000-2099/2066.Account Balance/Solution.sql",
+                "solution/2200-2299/2230.The Users That Are Eligible for Discount/Solution.sql",
                 "solution/2600-2699/2686.Immediate Food Delivery III/Solution.sql",
                 "solution/2700-2799/2752.Customers with Maximum Number of Transactions on Consecutive Days/Solution.sql"
             ],

solution/1900-1999/1910.Remove All Occurrences of a Substring/README.md

@@ -55,22 +55,76 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：暴力替换**
+
+我们循环判断 $s$ 中是否存在字符串 $part$，是则进行一次替换，继续循环此操作，直至 $s$ 中不存在字符串 $part$，返回此时的 $s$ 作为答案字符串。
+
+时间复杂度 $O(n^2 + n \times m)$，空间复杂度 $O(n + m)$。其中 $n$ 和 $m$ 分别是字符串 $s$ 和字符串 $part$ 的长度。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def removeOccurrences(self, s: str, part: str) -> str:
+        while part in s:
+            s = s.replace(part, '', 1)
+        return s
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public String removeOccurrences(String s, String part) {
+        while (s.contains(part)) {
+            s = s.replaceFirst(part, "");
+        }
+        return s;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    string removeOccurrences(string s, string part) {
+        int m = part.size();
+        while (s.find(part) != -1) {
+            s = s.erase(s.find(part), m);
+        }
+        return s;
+    }
+};
+```
+
+### **Go**
+
+```go
+func removeOccurrences(s string, part string) string {
+	for strings.Contains(s, part) {
+		s = strings.Replace(s, part, "", 1)
+	}
+	return s
+}
+```
+
+### **TypeScript**
 
+```ts
+function removeOccurrences(s: string, part: string): string {
+    while (s.includes(part)) {
+        s = s.replace(part, '');
+    }
+    return s;
+}
 ```
 
 ### **...**

solution/1900-1999/1910.Remove All Occurrences of a Substring/README_EN.md

@@ -56,13 +56,61 @@ Now s has no occurrences of "xy".
 ### **Python3**
 
 ```python
-
+class Solution:
+    def removeOccurrences(self, s: str, part: str) -> str:
+        while part in s:
+            s = s.replace(part, '', 1)
+        return s
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public String removeOccurrences(String s, String part) {
+        while (s.contains(part)) {
+            s = s.replaceFirst(part, "");
+        }
+        return s;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    string removeOccurrences(string s, string part) {
+        int m = part.size();
+        while (s.find(part) != -1) {
+            s = s.erase(s.find(part), m);
+        }
+        return s;
+    }
+};
+```
+
+### **Go**
+
+```go
+func removeOccurrences(s string, part string) string {
+	for strings.Contains(s, part) {
+		s = strings.Replace(s, part, "", 1)
+	}
+	return s
+}
+```
+
+### **TypeScript**
 
+```ts
+function removeOccurrences(s: string, part: string): string {
+    while (s.includes(part)) {
+        s = s.replace(part, '');
+    }
+    return s;
+}
 ```
 
 ### **...**

solution/1900-1999/1910.Remove All Occurrences of a Substring/Solution.cpp

@@ -0,0 +1,10 @@
+class Solution {
+public:
+    string removeOccurrences(string s, string part) {
+        int m = part.size();
+        while (s.find(part) != -1) {
+            s = s.erase(s.find(part), m);
+        }
+        return s;
+    }
+};

solution/1900-1999/1910.Remove All Occurrences of a Substring/Solution.go

@@ -0,0 +1,6 @@
+func removeOccurrences(s string, part string) string {
+	for strings.Contains(s, part) {
+		s = strings.Replace(s, part, "", 1)
+	}
+	return s
+}

solution/1900-1999/1910.Remove All Occurrences of a Substring/Solution.java

@@ -0,0 +1,8 @@
+class Solution {
+    public String removeOccurrences(String s, String part) {
+        while (s.contains(part)) {
+            s = s.replaceFirst(part, "");
+        }
+        return s;
+    }
+}

solution/1900-1999/1910.Remove All Occurrences of a Substring/Solution.py

@@ -0,0 +1,5 @@
+class Solution:
+    def removeOccurrences(self, s: str, part: str) -> str:
+        while part in s:
+            s = s.replace(part, '', 1)
+        return s

solution/1900-1999/1910.Remove All Occurrences of a Substring/Solution.ts

@@ -0,0 +1,6 @@
+function removeOccurrences(s: string, part: string): string {
+    while (s.includes(part)) {
+        s = s.replace(part, '');
+    }
+    return s;
+}

solution/1900-1999/1934.Confirmation Rate/README.md

@@ -98,7 +98,17 @@ Confirmations 表:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    user_id,
+    round(
+        sum(if(action = 'confirmed', 1, 0)) / count(1),
+        2
+    ) AS confirmation_rate
+FROM
+    Signups
+    LEFT JOIN Confirmations USING (user_id)
+GROUP BY user_id;
 ```
 
 <!-- tabs:end -->

solution/1900-1999/1934.Confirmation Rate/README_EN.md

@@ -94,7 +94,17 @@ User 2 made 2 requests where one was confirmed and the other timed out. The conf
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    user_id,
+    round(
+        sum(if(action = 'confirmed', 1, 0)) / count(1),
+        2
+    ) AS confirmation_rate
+FROM
+    Signups
+    LEFT JOIN Confirmations USING (user_id)
+GROUP BY user_id;
 ```
 
 <!-- tabs:end -->