Reduce-Based Map and Filter Pipeline
Reported by candidates from Tesla's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The mistake that sinks a first attempt on this Tesla OA, reported in September 2026, is returning the wrong shape when the filter wipes everything out. The problem is a simple map, filter, reduce pipeline over an int array, nothing exotic. But the output is three rows, and the third row must always be a one-element row, even when the filtered row is empty. Miss that and you fail the edge case. It's an easy problem with a few sharp corners. If you blank on the live assessment, StealthCoder runs invisibly as a safety net and hands you the clean version.
The problem
Run three functional stages on values: Map each value to value * multiplier. Filter the mapped sequence to values greater than or equal to minimum. Reduce the filtered sequence by summing it from an initial value of zero. Return three rows: the mapped sequence, the filtered sequence, and a one-element row containing the sum. Preserve order. Implementing reusable reduce-based helpers is encouraged. Function functionalStages(values: int[], multiplier: int, minimum: int) → int[][] Examples Example 1 values = [1,2,3] multiplier = 2 minimum = 4 return = [[2,4,6],[4,6],[10]] Mapping doubles the values; 4 and 6 survive and sum to 10. Example 2 values = [-2,0,3] multiplier = -1 minimum = 0 return = [[2,0,-3],[2,0],[2]] Order remains unchanged after mapping and filtering. Example 3 values = [1,2] multiplier = 3 minimum = 10 return = [[3,6],[],[0]] An empty filtered sequence reduces to the zero initial value. Constraints 0 <= values.length <= 10^5. All products and the final sum fit signed 32-bit range.
Reported by candidates. Source: FastPrep
Pattern and pitfall
There's no real hash table here, despite the hint. It's three linear passes, or one pass if you want. Map each value to value * multiplier, keep results that are >= minimum, then sum the kept ones starting from zero. Total cost is O(n) time and O(n) space. The traps are small but real. First, an empty filtered row must still produce [0] as the third row, as Example 3 shows. Second, keep the order exactly as given, so don't sort or dedupe. Third, filter the mapped values, not the originals. A negative multiplier flips signs, as Example 2 shows, so comparing the raw input to minimum gives wrong answers. Fourth, an empty input array should return [[],[],[0]]. The statement says products and the sum fit in 32 bits, so don't overthink overflow. If you freeze during the live OA, StealthCoder is the hedge that gives you a working solution fast.
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Reduce-Based Map and Filter Pipeline FAQ
How hard is the Tesla reduce-based map and filter problem really?+
Easy. It's three linear passes over an array. The difficulty is in the output format, not the algorithm. Most failures come from the empty-filter case or from filtering the original values instead of the mapped ones. If you read the examples carefully, you're fine.
What's the trick to this problem?+
There isn't a clever one. Map first, filter the mapped values, then sum with an initial value of zero. The reduce starting at zero is what makes an empty filtered row return [0] automatically. Write the reduce helper with an explicit initial value and the edge case handles itself.
What should I return when nothing passes the filter?+
Return the mapped row, an empty array for the filtered row, and [0] for the sum row. Example 3 shows exactly this: [[3,6],[],[0]]. The sum row always has exactly one element, regardless of what the filter keeps.
Do I need a hash table for this one?+
No. The hinted hash-table pattern doesn't apply, since nothing needs lookups or counting. A plain array pass is enough. Don't add a map or set, because that risks breaking order or dropping duplicates, and the problem says to preserve both.
How do I prepare for this in 48 hours?+
Practice writing map, filter, and reduce by hand in your language, including reduce with an initial value. Then test three cases: negative multiplier, empty input, and an everything-filtered-out case. That covers every trap in this problem, and it takes under an hour.