Human traffic of stadium

#
# 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.
#

# Write your MySQL query statement below

SELECT DISTINCT s1.*
FROM Stadium s1, Stadium s2, Stadium s3
WHERE s1.people >= 100 AND s2.people >= 100 AND s3.people >= 100
  AND (
  (s1.id + 1 = s2.id AND s2.id + 1 = s3.id) OR
  (s2.id + 1 = s1.id AND s1.id + 1 = s3.id) OR
  (s2.id + 1 = s3.id AND s3.id + 1 = s1.id)
  )
ORDER BY s1.visit_date;