How should I practice Cursor-Based Pagination Over Sorted Logs?

Start by reading the full statement, solve the visible examples, then compare the pattern against related Roblox onsite interview problems in FastPrep.

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Problem · Array

Cursor-Based Pagination Over Sorted Logs

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Problem statement

You are given logsByUser, where logsByUser[u] is the sorted list of log timestamps for user u.

A cursor-based paginator should return the next pageSize unread logs across all users in globally sorted order, then advance the cursor so the following call resumes from the next unread log.

For this FastPrep draft, implement the same cursor behavior as a single function: starting from the initial cursor, return every page that would be produced by repeatedly calling the paginator until no logs remain.

When two users have the same timestamp, emit the smaller user index first. The final page may contain fewer than pageSize logs. If logsByUser contains no logs or pageSize <= 0, return an empty array.

Function

paginateSortedLogs(logsByUser: int[][], pageSize: int) → int[][]

Examples

Example 1

logsByUser = [[1, 2, 9, 10, 11], [4, 5, 6, 7, 8]]pageSize = 3return = [[1, 2, 4], [5, 6, 7], [8, 9, 10], [11]]

The global sorted merge is [1, 2, 4, 5, 6, 7, 8, 9, 10, 11]. Repeated cursor calls with count 3 return the next slice each time.

Example 2

logsByUser = [[], [1, 2, 3], []]pageSize = 10return = [[1, 2, 3]]

Empty user log lists are skipped. Because pageSize is larger than the remaining logs, all logs appear in one page.

Example 3

logsByUser = [[1, 5], [1, 5]]pageSize = 4return = [[1, 1, 5, 5]]

Duplicate timestamps are allowed. Equal timestamps are emitted in stable user-index order.

Constraints

0 <= logsByUser.length
Each logsByUser[u] is sorted in nondecreasing order.
pageSize may be any integer.
The cursor state can be represented by the next unread index for each user.