# LeetCode Solution, Easy, 1729. Find Followers Count

# [1729. Find Followers Count](https://leetcode.com/problems/find-followers-count/)

## 題目敘述

SQL Schema >

    Create table If Not Exists Followers(user_id int, follower_id int)
    Truncate table Followers
    insert into Followers (user_id, follower_id) values ('0', '1')
    insert into Followers (user_id, follower_id) values ('1', '0')
    insert into Followers (user_id, follower_id) values ('2', '0')
    insert into Followers (user_id, follower_id) values ('2', '1')

Table: `Followers`

    +-------------+------+
    | Column Name | Type |
    +-------------+------+
    | user_id     | int  |
    | follower_id | int  |
    +-------------+------+
    (user_id, follower_id) is the primary key for this table.
    This table contains the IDs of a user and a follower in a social media app where the follower follows the user.

Write an SQL query that will, for each user, return the number of followers.

Return the result table ordered by `user_id` in ascending order.

The query result format is in the following example.

**Example 1:**

    Input:
    Followers table:
    +---------+-------------+
    | user_id | follower_id |
    +---------+-------------+
    | 0       | 1           |
    | 1       | 0           |
    | 2       | 0           |
    | 2       | 1           |
    +---------+-------------+
    Output:
    +---------+----------------+
    | user_id | followers_count|
    +---------+----------------+
    | 0       | 1              |
    | 1       | 1              |
    | 2       | 2              |
    +---------+----------------+
    Explanation:
    The followers of 0 are {1}
    The followers of 1 are {0}
    The followers of 2 are {0,1}

### 題目翻譯

找出每個 `user_id` 有多少不重複的 `follower_id`，並且最後按照 `user_id` 做升冪排序

## 解法解析

不重複 => DISTINCT, GROUP BY
升冪排序 => ORDER BY ASC

### 解法範例

#### MySQL

```sql
# Write your MySQL query statement below
SELECT user_id,
    COUNT(DISTINCT follower_id) AS followers_count
FROM Followers
GROUP BY user_id
ORDER BY user_id ASC
```