Create 584. Find Customer Referee

#2023/05/05
# Runtime Beats 46.15% if we use IFNULL

Author: Frisk0316

584. Find Customer Referee

@@ -0,0 +1,54 @@
+/*
+Problem Description: 
+
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| id          | int     |
+| name        | varchar |
+| referee_id  | int     |
++-------------+---------+
+id is the primary key column for this table.
+Each row of this table indicates the id of a customer, their name, and the id of the customer who referred them.
+
+Write an SQL query to report the names of the customer that are not referred by the customer with id = 2.
+
+Return the result table in any order.
+
+The query result format is in the following example.
+
+Example 1:
+
+Input: 
+Customer table:
++----+------+------------+
+| id | name | referee_id |
++----+------+------------+
+| 1  | Will | null       |
+| 2  | Jane | null       |
+| 3  | Alex | 2          |
+| 4  | Bill | null       |
+| 5  | Zack | 1          |
+| 6  | Mark | 2          |
++----+------+------------+
+Output: 
++------+
+| name |
++------+
+| Will |
+| Jane |
+| Bill |
+| Zack |
++------+
+*/
+
+# Write your MySQL query statement below
+
+SELECT name
+FROM Customer
+
+# COALESCE is used to replace NULL values with zero before checking whether it is equal to 2 or not.
+WHERE COALESCE(referee_id,0) <> 2;
+
+# Or you can use IFNULL to determine, which is faster
+WHERE IFNULL(referee_id,0) <> 2;