Customer Who Visited but Did Not Make Any Transactions | Leetcode 1581

Customer Who Visited but Did Not Make Any Transactions | Leetcode 1581 | Crack SQL Interviews

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Want to crack SQL interviews? Check out our latest video !!!
A 50-questions SQL study plan to ace any interview.
This tutorial will walk you through the solution in easy steps. This is an easy-level question from Leetcode.
Consistency is what transforms average into excellence.
So be consistent & keep Coding💻!!
SQL Study Plan : leetcode.com/s...
SQL Playlist: • Crack SQL Interview in...
Timecodes
0:00 - Introduction
0:26 - Question Explanation
2:30 - Understanding Concept with the help of Example
5:11 - Writing SQL Query
7:03 - Explanation of the Query
8:40 - Outro
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Пікірлер: 21

@learnwithchirag 8 ай бұрын

Few Test Cases were changed in the question after I made my video..... So if the above solution is giving a wrong answer when submitted then kindly change "t.transaction_id to t.amount" in the WHERE clause. So the new solution will be - ----------------------------------------------------------------------------------------------------------- SELECT v.customer_id, COUNT(v.visit_id) AS count_no_trans from Visits v LEFT JOIN Transactions t ON v.visit_id = t.visit_id WHERE t.amount IS NULL GROUP BY v.customer_id; -------------------------------------------------------------------------------------------------------------

@prathamsrivastava3153 7 ай бұрын

sir even the updated solution is not getting accepted..please check and help

@learnwithchirag 7 ай бұрын

@@prathamsrivastava3153 Hi pratham , They might have updated the test cases again. This can be done by sub queries method which I have taught later in this series- ---------------------------------------------------------------------------------------------------------------- SELECT customer_id, COUNT(visit_id) as count_no_trans FROM Visits WHERE visit_id NOT IN ( SELECT visit_id FROM Transactions ) GROUP BY customer_id ---------------------------------------------------------------------------------------------------------------- Or this way - ---------------------------------------------------------------------------------------------------------------- SELECT customer_id, COUNT(visit_id) as count_no_trans FROM Visits v WHERE NOT EXISTS ( SELECT visit_id FROM Transactions t WHERE t.visit_id = v.visit_id ) GROUP BY customer_id ---------------------------------------------------------------------------------------------------------------- WIll make a video on it in future whenever i get time ! To understand this solution you may complete the series once and then come back to this question again.

@prathamsrivastava3153 7 ай бұрын

Thank you so much sir@@learnwithchirag

@prathamsrivastava3153 6 ай бұрын

select customer_id,count(customer_id)as count_no_trans from Visits as v left join Transactions as t on v.visit_id=t.visit_id where transaction_id is null group by customer_id All cases accepted👍👍

@learnwithchirag 5 ай бұрын

Hey there! 👋 For more interesting content, tutorials, and updates, Feel free to connect with me on Instagram Handles :- @createwithchirag - instagram.com/createwithchirag/ @learn.with.chirag - instagram.com/learn.with.chirag/ LinkedIn: www.linkedin.com/in/chirag-sehgal-9200111b8/ Let's stay connected and keep the creativity flowing! 💡

@RaghunathGhate-e2c Ай бұрын

Started following you from yesterday and i feel that your videos are great and i am feeling confident now ...thank you so much and please make content related to big data.