1907. Count Salary Categories

LeetCode SQL 50 Solution/1907. Count Salary Categories.sql

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+907. Count Salary Categories
+Solved
+Medium
+Topics
+Companies
+SQL Schema
+Pandas Schema
+Table: Accounts
+
++-------------+------+
+| Column Name | Type |
++-------------+------+
+| account_id  | int  |
+| income      | int  |
++-------------+------+
+account_id is the primary key (column with unique values) for this table.
+Each row contains information about the monthly income for one bank account.
+
+
+Write a solution to calculate the number of bank accounts for each salary category. The salary categories are:
+
+"Low Salary": All the salaries strictly less than $20000.
+"Average Salary": All the salaries in the inclusive range [$20000, $50000].
+"High Salary": All the salaries strictly greater than $50000.
+The result table must contain all three categories. If there are no accounts in a category, return 0.
+
+Return the result table in any order.
+
+The result format is in the following example.
+
+
+
+Example 1:
+
+Input: 
+Accounts table:
++------------+--------+
+| account_id | income |
++------------+--------+
+| 3          | 108939 |
+| 2          | 12747  |
+| 8          | 87709  |
+| 6          | 91796  |
++------------+--------+
+Output: 
++----------------+----------------+
+| category       | accounts_count |
++----------------+----------------+
+| Low Salary     | 1              |
+| Average Salary | 0              |
+| High Salary    | 3              |
++----------------+----------------+
+Explanation: 
+Low Salary: Account 2.
+Average Salary: No accounts.
+High Salary: Accounts 3, 6, and 8.
+
+
+# Write your MySQL query statement below
+WITH
+    S AS (
+        SELECT 'Low Salary' AS category
+        UNION
+        SELECT 'Average Salary'
+        UNION
+        SELECT 'High Salary'
+    ),
+    T AS (
+        SELECT
+            CASE
+                WHEN income < 20000 THEN "Low Salary"
+                WHEN income > 50000 THEN 'High Salary'
+                ELSE 'Average Salary'
+            END AS category,
+            COUNT(1) AS accounts_count
+        FROM Accounts
+        GROUP BY 1
+    )
+SELECT category, IFNULL(accounts_count, 0) AS accounts_count
+FROM
+    S
+    LEFT JOIN T USING (category);
+
+
+
+### 1. Common Table Expression (CTE) "S"
+
+
+WITH
+    S AS (
+        SELECT 'Low Salary' AS category
+        UNION
+        SELECT 'Average Salary'
+        UNION
+        SELECT 'High Salary'
+    ),
+
+
+- **Purpose:**  
+  This CTE defines a static list of salary categories.  
+- **How it works:**  
+  - The `SELECT` statements with `UNION` combine three rows, each containing one of the categories: `'Low Salary'`, `'Average Salary'`, and `'High Salary'`.
+- **Result:**  
+  The result of `S` is a temporary table with one column (`category`) and three rows.
+
+---
+
+### 2. Common Table Expression (CTE) "T"
+
+```sql
+    T AS (
+        SELECT
+            CASE
+                WHEN income < 20000 THEN "Low Salary"
+                WHEN income > 50000 THEN 'High Salary'
+                ELSE 'Average Salary'
+            END AS category,
+            COUNT(1) AS accounts_count
+        FROM Accounts
+        GROUP BY 1
+    )
+```
+
+- **Purpose:**  
+  This CTE categorizes each account from the `Accounts` table based on the `income` value, then counts the number of accounts in each category.
+- **How it works:**  
+  - **CASE Statement:**  
+    - If `income` is less than 20000, it labels the row as `"Low Salary"`.
+    - If `income` is greater than 50000, it labels the row as `"High Salary"`.
+    - Otherwise, it labels the row as `"Average Salary"`.
+  - **COUNT(1):**  
+    - It counts the number of rows (accounts) in each category.
+  - **GROUP BY 1:**  
+    - It groups the results by the first column in the SELECT list, which is the computed `category`.
+- **Result:**  
+  The result of `T` is a temporary table that contains two columns: `category` and `accounts_count`. It holds the count of accounts for each salary category that exists in the `Accounts` table.
+
+---
+
+### 3. Final SELECT with LEFT JOIN
+
+```sql
+SELECT category, IFNULL(accounts_count, 0) AS accounts_count
+FROM
+    S
+    LEFT JOIN T USING (category);
+```
+
+- **Purpose:**  
+  This part combines the two CTEs (`S` and `T`) to ensure that every salary category from `S` is included in the final result, even if there are no corresponding accounts in `T`.
+- **How it works:**
+  - **LEFT JOIN:**  
+    - It joins `S` (all predefined categories) with `T` (the computed counts) on the `category` column.
+    - If a category from `S` does not exist in `T` (i.e., there were no accounts that fell into that category), the join will produce a `NULL` value for `accounts_count`.
+  - **IFNULL(accounts_count, 0):**  
+    - This function replaces any `NULL` in `accounts_count` with `0`, ensuring that the final output shows 0 for categories with no accounts.
+- **Result:**  
+  The final output is a list of salary categories along with the corresponding count of accounts. If a category has no accounts, it will show as 0.
+
+---
+
+### Summary
+
+- **CTE "S":** Defines a static list of salary categories.
+- **CTE "T":** Categorizes and counts accounts from the `Accounts` table based on income.
+- **LEFT JOIN:** Combines both CTEs so every predefined category appears in the final result, with missing counts defaulting to 0.
+- **Final Output:**  
+  A table with two columns:
+  - `category`: The salary category (Low Salary, Average Salary, High Salary).
+  - `accounts_count`: The number of accounts in that category (or 0 if there are none).