Leetcode #1336: Number of Transactions per Visit
In this guide, we solve Leetcode #1336 Number of Transactions per Visit in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Table: Visits +---------------+---------+ | Column Name | Type | +---------------+---------+ | user_id | int | | visit_date | date | +---------------+---------+ (user_id, visit_date) is the primary key (combination of columns with unique values) for this table. Each row of this table indicates that user_id has visited the bank in visit_date.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Database
Intuition
The task is relational in nature, which maps cleanly to DataFrame operations in Python.
By treating tables as DataFrames, joins and group-bys become concise and readable.
Approach
Load the inputs as DataFrames and apply the appropriate merge, filter, or group-by.
Select or rename the columns to match the required output.
Steps:
- Load inputs as DataFrames.
- Apply merge/groupby/filter operations.
- Select the output columns.
Example
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| user_id | int |
| visit_date | date |
+---------------+---------+
(user_id, visit_date) is the primary key (combination of columns with unique values) for this table.
Each row of this table indicates that user_id has visited the bank in visit_date.
Python Solution
import duckdb
import pandas as pd
def solution(visits: pd.DataFrame, transactions: pd.DataFrame) -> pd.DataFrame:
con = duckdb.connect()
con.register("Visits", visits)
con.register("Transactions", transactions)
return con.execute("""WITH RECURSIVE
S AS (
SELECT 0 AS n
UNION
SELECT n + 1
FROM S
WHERE
n < (
SELECT MAX(cnt)
FROM
(
SELECT COUNT(1) AS cnt
FROM Transactions
GROUP BY user_id, transaction_date
) AS t
)
),
T AS (
SELECT v.user_id, visit_date, IFNULL(cnt, 0) AS cnt
FROM
Visits AS v
LEFT JOIN (
SELECT user_id, transaction_date, COUNT(1) AS cnt
FROM Transactions
GROUP BY 1, 2
) AS t
ON v.user_id = t.user_id AND v.visit_date = t.transaction_date
)
SELECT n AS transactions_count, COUNT(user_id) AS visits_count
FROM
S AS s
LEFT JOIN T AS t ON s.n = t.cnt
GROUP BY n
ORDER BY n;""").df()
Complexity
The time complexity is O(n log n) (typical). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.