Last updated on October 9th, 2024 at 09:59 pm
This Leetcode problem Number of Transactions per Visit LeetCode Solution is done in SQL.
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Level of Question
Hard
Number of Transactions per Visit LeetCode Solution
Table of Contents
Problem Statement
Column Name | Type |
user_id | int |
visit_date | date |
(user_id, visit_date) is the primary key for this table.
Each row of this table indicates that user_id has visited the bank in visit_date.
Column Name | Type |
user_id | int |
transaction_date | date |
amount | int |
There is no primary key for this table, it may contain duplicates.
Each row of this table indicates that user_id has done a transaction of amount in transaction_date.
It is guaranteed that the user has visited the bank in the transaction_date.(i.e The Visits table contains (user_id, transaction_date) in one row)
A bank wants to draw a chart of the number of transactions bank visitors did in one visit to the bank and the corresponding number of visitors who have done this number of transaction in one visit.
Write an SQL query to find how many users visited the bank and didn’t do any transactions, how many visited the bank and did one transaction and so on.
The result table will contain two columns:
transactions_count which is the number of transactions done in one visit.
visits_count which is the corresponding number of users who did
transactions_count in one visit to the bank.
transactions_count should take all values from 0 to max(transactions_count) done by one or more users.
Order the result table by transactions_count.
The result format is in the following example.
Example 1:
Input:
user_id | visit_date |
1 | 2020-01-01 |
2 | 2020-01-02 |
12 | 2020-01-01 |
19 | 2020-01-03 |
1 | 2020-01-02 |
2 | 2020-01-03 |
1 | 2020-01-04 |
7 | 2020-01-11 |
9 | 2020-01-25 |
8 | 2020-01-28 |
user_id | transaction_date | amount |
1 | 2020-01-02 | 120 |
2 | 2020-01-03 | 22 |
7 | 2020-01-11 | 232 |
1 | 2020-01-04 | 7 |
9 | 2020-01-25 | 33 |
9 | 2020-01-25 | 66 |
8 | 2020-01-28 | 1 |
9 | 2020-01-25 | 99 |
Output:
transactions_count | visits_count |
0 | 4 |
1 | 5 |
2 | 0 |
3 | 1 |
Explanation:
– For transactions_count = 0, The visits (1, “2020-01-01”), (2, “2020-01-02”), (12, “2020-01-01”) and (19, “2020-01-03”) did no transactions so visits_count = 4.
– For transactions_count = 1, The visits (2, “2020-01-03”), (7, “2020-01-11”), (8, “2020-01-28”), (1, “2020-01-02”) and (1, “2020-01-04”) did one transaction so visits_count = 5.
– For transactions_count = 2, No customers visited the bank and did two transactions so visits_count = 0.
– For transactions_count = 3, The visit (9, “2020-01-25”) did three transactions so visits_count = 1.
– For transactions_count >= 4, No customers visited the bank and did more than three transactions so we will stop at transactions_count = 3
1. Number of Transactions per Visit LeetCode Solution MySQL
select ( select 0 ) as transactions_count, count(*) as visits_count from Visits where (user_id, visit_date) not in ( select user_id, transaction_date from Transactions ) union select s.transactions_count, if( visits_count is null, 0, visits_count ) as visits_count from ( select tc as transactions_count from ( select t.user_id, @tc := @tc + 1 as tc from Transactions as t, ( select @tc := 0 ) as u ) as s where tc <= ( select ifnull( max(transactions_count), 0 ) from ( select count(*) as transactions_count from Transactions group by user_id, transaction_date ) as t ) ) as s left join ( select transactions_count, count(*) as visits_count from ( select count(*) as transactions_count from Transactions group by user_id, transaction_date ) as t group by transactions_count ) as t on s.transactions_count = t.transactions_count order by transactions_count;