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601. Human Traffic of Stadium

Description

Table: Stadium

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| visit_date    | date    |
| people        | int     |
+---------------+---------+
visit_date is the column with unique values for this table.
Each row of this table contains the visit date and visit id to the stadium with the number of people during the visit.
As the id increases, the date increases as well.

 

Write a solution to display the records with three or more rows with consecutive id's, and the number of people is greater than or equal to 100 for each.

Return the result table ordered by visit_date in ascending order.

The result format is in the following example.

 

Example 1:

Input: 
Stadium table:
+------+------------+-----------+
| id   | visit_date | people    |
+------+------------+-----------+
| 1    | 2017-01-01 | 10        |
| 2    | 2017-01-02 | 109       |
| 3    | 2017-01-03 | 150       |
| 4    | 2017-01-04 | 99        |
| 5    | 2017-01-05 | 145       |
| 6    | 2017-01-06 | 1455      |
| 7    | 2017-01-07 | 199       |
| 8    | 2017-01-09 | 188       |
+------+------------+-----------+
Output: 
+------+------------+-----------+
| id   | visit_date | people    |
+------+------------+-----------+
| 5    | 2017-01-05 | 145       |
| 6    | 2017-01-06 | 1455      |
| 7    | 2017-01-07 | 199       |
| 8    | 2017-01-09 | 188       |
+------+------------+-----------+
Explanation: 
The four rows with ids 5, 6, 7, and 8 have consecutive ids and each of them has >= 100 people attended. Note that row 8 was included even though the visit_date was not the next day after row 7.
The rows with ids 2 and 3 are not included because we need at least three consecutive ids.

Solutions

Solution 1

Thinking

We need at least three consecutive days with people >= 100. A 3-way self-join covers windows of length exactly \(3\), but longer streaks become awkward.

After filtering qualifying rows, consecutive ids share a constant id - ROW_NUMBER(). Group by that difference, count each group, and keep groups of size at least \(3\).

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# Write your MySQL query statement below
WITH
    S AS (
        SELECT
            *,
            id - (ROW_NUMBER() OVER (ORDER BY id)) AS rk
        FROM Stadium
        WHERE people >= 100
    ),
    T AS (SELECT *, COUNT(1) OVER (PARTITION BY rk) AS cnt FROM S)
SELECT id, visit_date, people
FROM T
WHERE cnt >= 3
ORDER BY 1;

Solution 2

Thinking

Method 1 already counts group size with a window. The same difference id_diff can be aggregated with HAVING COUNT(*) > 2 and used as an IN filter. The meaning is unchanged; only the counting style differs.

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# Write your MySQL query statement below
WITH
    Consecutive AS (
        SELECT
            *,
            id - ROW_NUMBER() OVER () AS id_diff
        FROM Stadium
        WHERE people >= 100
    )
SELECT id, visit_date, people
FROM Consecutive
WHERE
    id_diff IN (
        SELECT id_diff
        FROM Consecutive
        GROUP BY id_diff
        HAVING COUNT(*) > 2
    )
ORDER BY visit_date;

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