Skip to content

Latest commit

 

History

History
192 lines (152 loc) · 5.47 KB

README_EN.md

File metadata and controls

192 lines (152 loc) · 5.47 KB
comments difficulty edit_url tags
true
Medium
Database

3124. Find Longest Calls 🔒

中文文档

Description

Table: Contacts

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| first_name  | varchar |
| last_name   | varchar |
+-------------+---------+
id is the primary key (column with unique values) of this table.
id is a foreign key (reference column) to Calls table.
Each row of this table contains id, first_name, and last_name.

Table: Calls

+-------------+------+
| Column Name | Type |
+-------------+------+
| contact_id  | int  |
| type        | enum |
| duration    | int  |
+-------------+------+
(contact_id, type, duration) is the primary key (column with unique values) of this table.
type is an ENUM (category) type of ('incoming', 'outgoing').
Each row of this table contains information about calls, comprising of contact_id, type, and duration in seconds.

Write a solution to find the three longest incoming and outgoing calls.

Return the result table ordered by type, duration, and first_name in descending order and duration must be formatted as HH:MM:SS.

The result format is in the following example.

 

Example 1:

Input:

Contacts table:

+----+------------+-----------+
| id | first_name | last_name |
+----+------------+-----------+
| 1  | John       | Doe       |
| 2  | Jane       | Smith     |
| 3  | Alice      | Johnson   |
| 4  | Michael    | Brown     |
| 5  | Emily      | Davis     |
+----+------------+-----------+

Calls table:

+------------+----------+----------+
| contact_id | type     | duration |
+------------+----------+----------+
| 1          | incoming | 120      |
| 1          | outgoing | 180      |
| 2          | incoming | 300      |
| 2          | outgoing | 240      |
| 3          | incoming | 150      |
| 3          | outgoing | 360      |
| 4          | incoming | 420      |
| 4          | outgoing | 200      |
| 5          | incoming | 180      |
| 5          | outgoing | 280      |
+------------+----------+----------+

Output:

+-----------+----------+-------------------+
| first_name| type     | duration_formatted|
+-----------+----------+-------------------+
| Michael   | incoming | 00:07:00          |
| Jane      | incoming | 00:05:00          |
| Emily     | incoming | 00:03:00          |
| Alice     | outgoing | 00:06:00          |
| Emily     | outgoing | 00:04:40          |
| Jane      | outgoing | 00:04:00          |
+-----------+----------+-------------------+

Explanation:

  • Michael had an incoming call lasting 7 minutes.
  • Jane had an incoming call lasting 5 minutes.
  • Emily had an incoming call lasting 3 minutes.
  • Alice had an outgoing call lasting 6 minutes.
  • Emily had an outgoing call lasting 4 minutes and 40 seconds.
  • Jane had an outgoing call lasting 4 minutes.

Note: Output table is sorted by type, duration, and first_name in descending order.

Solutions

Solution 1: Equi-Join + Window Function

We can use equi-join to connect the two tables, and then use the window function RANK() to calculate the ranking of each type of phone. Finally, we just need to filter out the top three phones.

MySQL

WITH
    T AS (
        SELECT
            first_name,
            type,
            DATE_FORMAT(SEC_TO_TIME(duration), "%H:%i:%s") AS duration_formatted,
            RANK() OVER (
                PARTITION BY type
                ORDER BY duration DESC
            ) AS rk
        FROM
            Calls AS c1
            JOIN Contacts AS c2 ON c1.contact_id = c2.id
    )
SELECT
    first_name,
    type,
    duration_formatted
FROM T
WHERE rk <= 3
ORDER BY 2, 3 DESC, 1 DESC;

Python3

import pandas as pd


def find_longest_calls(contacts: pd.DataFrame, calls: pd.DataFrame) -> pd.DataFrame:
    merged_data = calls.merge(contacts, left_on="contact_id", right_on="id")
    merged_data["duration_formatted"] = (
        merged_data["duration"] // 3600 * 10000
        + merged_data["duration"] % 3600 // 60 * 100
        + merged_data["duration"] % 60
    ).apply(lambda x: "{:02}:{:02}:{:02}".format(x // 10000, x // 100 % 100, x % 100))

    merged_data["rk"] = merged_data.groupby("type")["duration"].rank(
        method="dense", ascending=False
    )

    result = merged_data[merged_data["rk"] <= 3][
        ["first_name", "type", "duration_formatted"]
    ]
    result = result.sort_values(
        by=["type", "duration_formatted", "first_name"], ascending=[True, False, False]
    )
    return result