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Daily Leads and Partners

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].

Solution Explanation for LeetCode 1693: Daily Leads and Partners

This problem requires us to analyze a DailySales table and determine the number of unique leads and partners for each combination of date_id and make_name. The solution leverages the power of SQL's GROUP BY and COUNT(DISTINCT) clauses.

Approach

The core idea is to group the DailySales data by date_id and make_name. This ensures that we're analyzing each day's sales for each make separately. Then, for each group, we use COUNT(DISTINCT lead_id) to find the number of unique leads and COUNT(DISTINCT partner_id) to count unique partners.

SQL Solution (MySQL)

The provided MySQL query efficiently implements this approach:

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;
  • SELECT date_id, make_name, ...: This selects the date_id and make_name for each group, along with the calculated counts.
  • COUNT(DISTINCT lead_id) AS unique_leads: Counts the number of unique lead_id values within each group and names the resulting column unique_leads.
  • COUNT(DISTINCT partner_id) AS unique_partners: Similarly, counts the unique partner_id values and names the column unique_partners.
  • FROM DailySales: Specifies the table to query.
  • GROUP BY 1, 2: Groups the results by the first and second columns in the SELECT statement (i.e., date_id and make_name). This is a shorthand for GROUP BY date_id, make_name.

Time Complexity Analysis

The time complexity of this SQL query depends on the database engine's implementation of GROUP BY and COUNT(DISTINCT). However, it's generally considered to be O(N log N) or potentially O(N) in optimized scenarios, where N is the number of rows in the DailySales table. The GROUP BY operation requires sorting or hashing, leading to the logarithmic factor in the worst case. COUNT(DISTINCT) also incurs some computational overhead for distinct counting.

The space complexity is primarily determined by the size of the intermediate result set produced by the GROUP BY operation. In the worst case (all distinct combinations of date_id and make_name), it could be proportional to the number of unique combinations, but in practice, it will be less than or equal to the number of rows in the input table. Therefore, the space complexity is considered O(N) in the worst case.

Note: The exact time and space complexities can vary based on database indexing, query optimization techniques employed by the database engine, and the size and characteristics of the data.