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Q1693LeadsAndPartners.sql
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53 lines (45 loc) · 2.05 KB
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/*
1693. Daily Leads and Partners
Question description:
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.
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].
*/
/*Solution*/
select distinct 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;
/*
Explanation:
Use count to count number of unique lead_id and partner_id together with the keyword DISTINCT
Use group by to combine results with same date_id and make_name
*/