Question

1693. Daily Leads and Partners

中文文档

Description

Table: DailySales

+-------------+---------+
| Column Name | Type  |
+-------------+---------+
| date_id   | date  |
| make_name   | varchar |
| lead_id   | int   |
| partner_id  | int   |
+-------------+---------+
There is no primary key (column with unique values) for this table. It may contain duplicates.
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.

 

For each date_id and make_name, find the number of distinct lead_id's and distinct partner_id's.

Return the result table in any order.

The 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].

Solutions

Solution 1: Group By + Count Distinct

We can use the GROUP BY statement to group the data by the date_id and make_name fields, and then use the COUNT(DISTINCT) function to count the number of distinct values for lead_id and partner_id.

# 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 1, 2;