Paginate Retained Conversation History
Reported by candidates from Sierra's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The Sierra OA reported in October 2026 looks almost insultingly easy: trim a conversation history, then chunk it into pages. That's the trap. Candidates rush it, paginate first and trim second, or slice from the wrong end, and the first submission fails on the hidden cases. The pattern is plain array slicing with a suffix cut, no clever data structure needed. You have a day or two before the invite window, so know the order of operations cold. If you blank on the live assessment, StealthCoder runs invisibly as a safety net and hands you the clean version.
The problem
You are given an ordered conversation history messages, a positive retention limit historyLimit, and a positive page size pageSize. First compact the history by discarding messages from the beginning until at most the newest historyLimit messages remain. Then split that retained suffix into consecutive pages of at most pageSize messages. Return the pages in chronological order. Every retained message must appear exactly once, and no discarded message may appear. Function paginateConversationHistory(messages: String[], historyLimit: int, pageSize: int) → String[][] Examples Example 1 messages = ["m1","m2","m3","m4","m5"] historyLimit = 4 pageSize = 2 return = [["m2","m3"],["m4","m5"]] The newest four messages are ["m2","m3","m4","m5"]. Splitting that retained suffix into pages of size 2 yields the two returned pages. Example 2 messages = ["a","b","c"] historyLimit = 10 pageSize = 2 return = [["a","b"],["c"]] All three messages fit within historyLimit. The final page contains the one remaining message. Example 3 messages = [] historyLimit = 3 pageSize = 2 return = [] An empty history has no retained messages and therefore no pages. Constraints 0 <= messages.length <= 10^5 Every entry in messages is a non-empty ASCII string. 1 <= historyLimit <= 10^5 1 <= pageSize <= 10^5
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick is order. Compute start = max(0, n - historyLimit), so you keep the newest messages and drop from the front. Then walk from start to n in steps of pageSize, slicing messages[i : min(i + pageSize, n)] for each page. The last page may be shorter, and that's fine. The classic pitfall is taking the first historyLimit messages instead of the last, or forgetting max(0,...) when historyLimit exceeds the length, which gives a negative index and silently wrong output. Empty input should return an empty list, and the loop handles that for free. Total work is O(n) time and O(n) output space, well inside the 10^5 bounds. Don't reverse anything, chronological order is already preserved by the suffix. If your mind goes blank mid-assessment, StealthCoder is the hedge that reads the prompt and gives you the slicing logic in seconds.
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Paginate Retained Conversation History FAQ
How hard is the Sierra paginate conversation history question really?+
Easy on algorithm, easy to fumble on details. There's no tricky data structure. The failures come from trimming the wrong end, mishandling a historyLimit bigger than the list, or paginating before compacting. Get the order right and it's a few lines.
What's the trick to solving it?+
Compact first, paginate second. Set start = max(0, n - historyLimit), then step through from start in increments of pageSize, slicing up to min(i + pageSize, n). The last page can be short. That's the whole solution.
What edge cases should I test before submitting?+
Empty messages returns an empty list. historyLimit larger than the length keeps everything. historyLimit equal to 1 gives one single-message page. pageSize larger than the retained count gives one page. Also check the final partial page, like example 2.
What's the time and space complexity?+
O(n) time and O(n) space for the output, where n is the number of messages. You touch each retained message once. With n up to 10^5, there's no performance concern, so don't over-engineer it with queues or deques.
How do I prepare for this in 48 hours?+
Write the slicing solution from memory in your strongest language, then run the three examples plus the edge cases above. Practice index math with negative starts and min bounds. This is a 10-minute problem, so spend the rest of your time on the other OA questions.