countChallenges.txt

/*
Enter your query here.

Julia asked her students to create some coding challenges. Write a query to print the hacker_id, name, and the total number of challenges created by each student. Sort your results by the total number of challenges in descending order. If more than one student created the same number of challenges, then sort the result by hacker_id. If more than one student created the same number of challenges and the count is less than the maximum number of challenges created, then exclude those students from the result.

Hackers: The hacker_id is the id of the hacker, and name is the name of the hacker. 

Challenges: The challenge_id is the id of the challenge, and hacker_id is the id of the student who created the challenge. 



*/

SELECT h.hacker_id, h.name, COUNT(c.hacker_id) AS c_count
FROM Hackers AS h
JOIN Challenges AS c
ON h.hacker_id = c.hacker_id
GROUP BY c.hacker_id, h.name
HAVING c_count =
(SELECT MAX(temp1.cnt) FROM
    (SELECT COUNT(hacker_id) AS cnt 
    FROM Challenges AS c1
    GROUP BY c1.hacker_id ASC
    ORDER BY c1.hacker_id DESC) AS temp1)
OR c_count IN 
        (SELECT t.cnt
         FROM (SELECT COUNT(*) AS cnt 
               FROM Challenges
               GROUP BY hacker_id) AS t
         GROUP BY t.cnt
         HAVING COUNT(t.cnt) = 1)
ORDER BY c_count DESC, c.hacker_id;