# LeetCode Solution, Easy, 1693. Daily Leads and Partners

# [1693. Daily Leads and Partners](https://leetcode.com/problems/daily-leads-and-partners/)

## 題目敘述

SQL Schema >

    Create table If Not Exists DailySales(date_id date, make_name varchar(20), lead_id int, partner_id int)
    Truncate table DailySales
    insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-8', 'toyota', '0', '1')
    insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-8', 'toyota', '1', '0')
    insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-8', 'toyota', '1', '2')
    insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-7', 'toyota', '0', '2')
    insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-7', 'toyota', '0', '1')
    insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-8', 'honda', '1', '2')
    insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-8', 'honda', '2', '1')
    insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-7', 'honda', '0', '1')
    insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-7', 'honda', '1', '2')
    insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-7', 'honda', '2', '1')

Table: `DailySales`

    +-------------+---------+
    | Column Name | Type    |
    +-------------+---------+
    | date_id     | date    |
    | make_name   | varchar |
    | lead_id     | int     |
    | partner_id  | int     |
    +-------------+---------+
    This table does not have a primary key.
    This table contains the date and the name of the product sold and the IDs of the lead and partner it was sold to.
    The name consists of only lowercase English letters.

Write an SQL query that will, for each `date_id` and `make_name`, return the number of distinct `lead_id`'s and distinct `partner_id`'s.

Return the result table in any order.

The query result format is in the following example.

**Example 1:**

    Input:
    DailySales table:
    +-----------+-----------+---------+------------+
    | date_id   | make_name | lead_id | partner_id |
    +-----------+-----------+---------+------------+
    | 2020-12-8 | toyota    | 0       | 1          |
    | 2020-12-8 | toyota    | 1       | 0          |
    | 2020-12-8 | toyota    | 1       | 2          |
    | 2020-12-7 | toyota    | 0       | 2          |
    | 2020-12-7 | toyota    | 0       | 1          |
    | 2020-12-8 | honda     | 1       | 2          |
    | 2020-12-8 | honda     | 2       | 1          |
    | 2020-12-7 | honda     | 0       | 1          |
    | 2020-12-7 | honda     | 1       | 2          |
    | 2020-12-7 | honda     | 2       | 1          |
    +-----------+-----------+---------+------------+
    Output:
    +-----------+-----------+--------------+-----------------+
    | date_id   | make_name | unique_leads | unique_partners |
    +-----------+-----------+--------------+-----------------+
    | 2020-12-8 | toyota    | 2            | 3               |
    | 2020-12-7 | toyota    | 1            | 2               |
    | 2020-12-8 | honda     | 2            | 2               |
    | 2020-12-7 | honda     | 3            | 2               |
    +-----------+-----------+--------------+-----------------+
    Explanation:
    For 2020-12-8, toyota gets leads = [0, 1] and partners = [0, 1, 2] while honda gets leads = [1, 2] and partners = [1, 2].
    For 2020-12-7, toyota gets leads = [0] and partners = [1, 2] while honda gets leads = [0, 1, 2] and partners = [1, 2].

### 題目翻譯

要從資料表 `DailySales` 中取出每個 `date_id` 和 `make_name` 的組合中，不重複的 `lead_id` 和 `partner_id` 總數。

## 解法解析

光看到不重複的第一直覺就是要使用到 `DISTINCT`，然後再來是有組合的，就會需要搭配使用 `GROUP BY` 了。最後因為要算個數所以再搭配使用 `COUNT` 來做計算

### 解法範例

#### MySQL

```sql
# Write your MySQL query statement below
SELECT date_id,
    make_name,
    COUNT(DISTINCT lead_id) AS unique_leads,
    COUNT(DISTINCT partner_id) AS unique_partners
FROM DailySales
GROUP BY date_id,
    make_name
```