Is Create a Session Bar Chart easy or hard?

Create a Session Bar Chart is classified as an Easy database problem with a high acceptance rate around 75%. The main requirement is understanding how to categorize values into ranges and count them using SQL aggregation.

Create a Session Bar Chart Python/Java solution

This problem is a SQL database query rather than a typical algorithm implemented in Python or Java. The solution is written in MySQL using CASE expressions, GROUP BY, or conditional aggregation to compute the counts for each duration range.

How to solve Create a Session Bar Chart in O(n)?

Perform a single pass over the Sessions table and classify durations into bins using CASE or conditional aggregation. Each row contributes to exactly one bucket, and SQL aggregation functions such as COUNT or SUM compute totals. Because the table is scanned once, the overall complexity remains O(n).

What is the best approach for Create a Session Bar Chart?

The most common solution uses a CASE expression to convert session durations into buckets and then aggregates them with GROUP BY. This performs a single table scan with O(n) time complexity and constant space. Conditional aggregation using SUM with boolean expressions is another equally efficient approach.

Is Create a Session Bar Chart asked at Google/Amazon/Meta?

Histogram-style SQL questions that group values into ranges appear frequently in interviews at large tech companies. Variations of this problem test your understanding of CASE expressions, aggregation, and analytical reporting queries used in real production databases.

What data structure is used in Create a Session Bar Chart?

The problem relies on SQL aggregation rather than traditional data structures. The database engine maintains counters for each duration bucket during the table scan, similar to how a histogram stores counts for predefined ranges.

What is the time complexity of Create a Session Bar Chart?

The query typically scans the Sessions table once to categorize durations and count rows. This results in O(n) time complexity where n is the number of sessions. Since aggregation uses only a few counters, the extra space complexity is O(1).

1435. Create a Session Bar Chart - Practice Online

Table: Sessions

+---------------------+---------+
| Column Name         | Type    |
+---------------------+---------+
| session_id          | int     |
| duration            | int     |
+---------------------+---------+
session_id is the column of unique values for this table.
duration is the time in seconds that a user has visited the application.

You want to know how long a user visits your application. You decided to create bins of "[0-5>", "[5-10>", "[10-15>", and "15 minutes or more" and count the number of sessions on it.

Write a solution to report the (bin, total).

Return the result table in any order.

The result format is in the following example.

Example 1:

Input: 
Sessions table:
+-------------+---------------+
| session_id  | duration      |
+-------------+---------------+
| 1           | 30            |
| 2           | 199           |
| 3           | 299           |
| 4           | 580           |
| 5           | 1000          |
+-------------+---------------+
Output: 
+--------------+--------------+
| bin          | total        |
+--------------+--------------+
| [0-5>        | 3            |
| [5-10>       | 1            |
| [10-15>      | 0            |
| 15 or more   | 1            |
+--------------+--------------+
Explanation: 
For session_id 1, 2, and 3 have a duration greater or equal than 0 minutes and less than 5 minutes.
For session_id 4 has a duration greater or equal than 5 minutes and less than 10 minutes.
There is no session with a duration greater than or equal to 10 minutes and less than 15 minutes.
For session_id 5 has a duration greater than or equal to 15 minutes.