MEDIUM

Author: evanmathew

1907. Count Salary Categories.sql

@@ -1,3 +1,61 @@
+-- Problem:
+
+/*
+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.
+*/
+
+-------------------------------------------------------------------------------
+
+-- Solution:
+
 SELECT "Low Salary" AS category , SUM(IF(income<20000,1,0)) AS accounts_count  FROM Accounts 
 UNION
 SELECT "Average Salary" AS category , SUM(IF(income>=20000 AND income<=50000,1,0)) AS accounts_count FROM Accounts