Ordered Payload Release
Reported by candidates from Arcesium's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
Arcesium reported this one in July 2026, and it looks fancier than it is. Strip the story and it's a reorder buffer: a hash map of held payloads plus a pointer to the next expected sequence number. If you've got an Arcesium OA coming, expect to write this loop cleanly and fast. The traps are duplicates and the per-arrival output rows. StealthCoder sits invisibly on your screen as a safety net if you blank on the live OA, but this pattern is easy to lock in before then.
The problem
Payloads arrive one at a time. The ith arrival has positive sequence number sequenceNumbers[i] and string payload payloads[i]. Payloads must be released in increasing sequence-number order, beginning with sequence number 1. Process the arrivals in their given order: If the arrival is ahead of the next expected sequence number, buffer it. After each arrival, immediately release every contiguous buffered payload beginning with the next expected sequence number. Release those payloads in increasing sequence-number order. If a sequence number has already been released, ignore a later arrival with that sequence number. Return one row for every arrival. Row i contains exactly the payloads released immediately after arrival i. Function releasePayloadsInOrder(sequenceNumbers: int[], payloads: String[]) → String[][] Examples Example 1 sequenceNumbers = [2,1,4,3,2,5] payloads = ["B","A","D","C","B","E"] return = [[],["A","B"],[],["C","D"],[],["E"]] Sequence 2 is buffered until sequence 1 arrives, so the second arrival releases A and B. The same pattern releases C and D when sequence 3 arrives. The later duplicate of sequence 2 is ignored because it was already released.
Reported by candidates. Source: FastPrep
Pattern and pitfall
The problem reduces to a hash map and a counter. Keep next = 1 and a map from sequence number to payload. For each arrival, if its sequence number is below next, it's already released, so ignore it. Otherwise store it in the map. Then while next is in the map, append that payload to the current row, remove it, and increment next. Push the row, even when it's empty, because you return exactly one row per arrival. The common pitfalls are forgetting the empty rows, releasing out of order, and mishandling duplicates that arrive while still buffered. A buffered duplicate (same number, not yet released) can just overwrite or be skipped, since the example only defines ignoring already-released numbers. Total work is O(n) because each payload is released once. If you freeze mid-assessment, StealthCoder can hand you this loop live, but you should be able to write it from memory.
Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.
You can drill Ordered Payload Release cold, or you can hedge it. StealthCoder runs invisibly during screen share and surfaces a working solution in under 2 seconds. The proctor sees the IDE. They don't see what's behind it. Made for the candidate who got the OA invite this morning and has 72 hours, not six months.
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Arcesium reuses patterns across OAs. Made for the candidate who got the OA invite this morning and has 72 hours, not six months. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Ordered Payload Release FAQ
How hard is Ordered Payload Release really?+
Easy to medium. There's no clever algorithm, just careful bookkeeping. A hash map and a next-expected pointer solve it in linear time. Most people lose points on edge cases like empty rows and duplicates, not on the core idea.
What's the trick to this Arcesium problem?+
Keep a pointer for the next expected sequence number and a map of buffered payloads. After every arrival, loop while the pointer exists in the map, release it, and advance. That drain loop is the entire solution.
How do I handle duplicate sequence numbers?+
If the number is less than the next expected one, it's already released, so skip it. If it's still buffered, storing it again in the map doesn't change the outcome. Either way, the row for that arrival may just be empty.
Do I return a row for arrivals that release nothing?+
Yes. The output has exactly one row per arrival, so buffered or ignored arrivals produce an empty list. Missing those rows shifts every later row and fails the tests.
How do I prepare for this in 48 hours?+
Write the reorder buffer from scratch twice with a map and a pointer. Then test it on the example plus cases with gaps, repeats of released numbers, and arrivals in perfectly sorted order. It takes under an hour to get comfortable.