Timestamped Telemetry Stream Differences
Reported by candidates from SpaceX's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
SpaceX reported this one in September 2026, and it looks almost too easy. Arrays go up to 10^5 elements, so anything quadratic is dead on arrival, but nothing here needs it. The task is a single aligned pass: subtract right from left at each index, keep the timestamp, emit a row. If you're taking this OA soon, the risk isn't the algorithm. It's overthinking it or fumbling the output shape. StealthCoder sits invisibly on your screen as a safety net if you blank mid-assessment, but this one you can likely write from memory.
The problem
Two telemetry streams contain aligned integer samples. Only the first stream supplies timestamps, in nondecreasing order. Arrays leftValues and rightValues contain the aligned measurements for those timestamps. Return one row [timestamp, leftValue - rightValue] for every sample, preserving input order. Function telemetryDifferences(timestamps: int[], leftValues: int[], rightValues: int[]) → int[][] Examples Example 1 timestamps = [10,20,30] leftValues = [7,4,9] rightValues = [2,6,3] return = [[10,5],[20,-2],[30,6]] Subtract aligned values and retain the timestamp from the first stream. Example 2 timestamps = [5] leftValues = [-3] rightValues = [4] return = [[5,-7]] Negative differences are retained. Example 3 timestamps = [1,1,2] leftValues = [5,8,6] rightValues = [5,3,9] return = [[1,0],[1,5],[2,-3]] Nondecreasing timestamps may repeat and every aligned sample is emitted. Constraints 1 <= timestamps.length <= 10^5. All three arrays have the same length. Timestamps are nondecreasing and every difference fits a signed 32-bit integer.
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick is that there's no trick. The three arrays are index-aligned, so row i is [timestamps[i], leftValues[i] - rightValues[i]]. One loop, O(n) time, O(n) output. The input size of 10^5 only rules out things like nested loops, searching for matching timestamps, or merging streams by value. Don't do any of that. The pitfall is reading the repeated timestamps in Example 3 as something to dedupe or group. Don't. Every aligned sample gets emitted, duplicates included, in original order. Second pitfall: sorting. The timestamps are already nondecreasing, so sorting only risks breaking alignment. Third, watch negative results and make sure you return pairs, not flat values. The problem says differences fit a signed 32-bit integer, so no overflow worry in most languages. If the format trips you up live, StealthCoder is the hedge, but it's a five-line function.
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Timestamped Telemetry Stream Differences FAQ
How hard is the SpaceX Timestamped Telemetry Stream Differences problem really?+
Very easy. It's a single pass over three equal-length arrays. The difficulty is only in reading carefully and returning the right shape, a list of [timestamp, difference] pairs. If you've written a basic loop, you can solve it in a couple of minutes.
What's the trick to this problem?+
There isn't one. Index i in all three arrays refers to the same sample. Compute leftValues[i] minus rightValues[i], pair it with timestamps[i], and append. Don't search, merge, or sort. The alignment already does the work for you.
Do I need to handle duplicate timestamps?+
No special handling. Example 3 shows timestamps repeating, and every aligned sample still gets its own row. Don't dedupe, group, or aggregate. Just emit one row per index in the original order.
What's the time complexity I should aim for?+
O(n) time and O(n) space for the output. With n up to 10^5, a linear pass is comfortable. Anything that compares samples against each other, like nested loops or lookups by timestamp, is unnecessary and could be too slow.
How should I prepare in 48 hours for an OA like this?+
Practice array iteration and output formatting in your chosen language, especially building lists of pairs. Then rehearse a few harder patterns, since other questions in the same OA may not be this simple. Check edge cases: a single element, negative differences, and repeated timestamps.