Challenges SQL Hacker Rank Solution

Hello coders, In this post, you will learn how to solve the Challenges SQL Hacker Rank Solution. This problem is a part of the SQL Hacker Rank series.

Problem

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.

Input Format

The following tables contain challenge data:

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.

Sample Input 0

Hackers Table: Challenges SQL Hacker Rank Solution Challenges Table:

Sample Output 0

21283 Angela 6
88255 Patrick 5
96196 Lisa 1

Sample Input 1

Hackers Table: Challenges SQL Hacker Rank Solution Challenges Table:

Sample Output 1

12299 Rose 6
34856 Angela 6
79345 Frank 4
80491 Patrick 3
81041 Lisa 1

Explanation

For Sample Case 0, we can get the following details:
Challenges SQL Hacker Rank Solution
Students and both created challenges, but the maximum number of challenges created is so these students are excluded from the result.

For Sample Case 1, we can get the following details:
Challenges SQL Hacker Rank Solution
Students and both created challenges. Because is the maximum number of challenges created, these students are included in the result.

Challenges SQL Hacker Rank Solution

SELECT H.HACKER_ID,
       H.NAME,
       COUNT(C.CHALLENGE_ID) AS TOTAL
FROM HACKERS H,
     CHALLENGES C
WHERE H.HACKER_ID=C.HACKER_ID
GROUP BY H.HACKER_ID,
         H.NAME
HAVING COUNT(C.CHALLENGE_ID) IN
  (SELECT MAX(TOTAL)
   FROM
     (SELECT COUNT(*) AS TOTAL
      FROM CHALLENGES
      GROUP BY HACKER_ID))
OR COUNT(C.CHALLENGE_ID) IN
  (SELECT TOTAL
   FROM
     (SELECT COUNT(*) AS TOTAL
      FROM CHALLENGES
      GROUP BY HACKER_ID)
   GROUP BY TOTAL
   HAVING COUNT(TOTAL)=1)
ORDER BY COUNT(C.CHALLENGE_ID) DESC, H.HACKER_ID;

Disclaimer: The above Problem (Challenges) generated by Hackerrank but the Solution is Provided by Chase2Learn. This tutorial is only for Educational and Learning purposes.