Most Frequent Reduced Digit
Reported by candidates from Airbnb's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The Airbnb OA reported in July 2026 dresses up a digital root problem as sensor data. Strip the story and it's two steps: collapse every number to a single digit, then count which digit shows up most. Ties go to the higher digit. That's it. If you've got an invite and 48 hours, this one is quick to nail down. The only real risk is overcomplicating it or fumbling the tie rule under pressure. StealthCoder sits invisibly on your screen during the live assessment as a safety net if you blank, but you probably won't need it once you see the shape.
The problem
You are working with data collected from various sensors. Given an array of non-negative integers readings, repeatedly replace each value with the sum of its decimal digits until every value is a single digit. Return the most frequent digit in the final transformed array. If several digits have the same highest frequency, return the highest such digit. A solution with time complexity no worse than O(readings.length^2) fits within the execution time limit. Function mostFrequentReducedDigit(readings: int[]) → int Examples Example 1 readings = [123,456,789,101] return = 6 123 becomes 1 + 2 + 3 = 6. 456 becomes 4 + 5 + 6 = 15, then 1 + 5 = 6. 789 becomes 7 + 8 + 9 = 24, then 2 + 4 = 6. 101 becomes 1 + 0 + 1 = 2. The final array is [6,6,6,2], so the most frequent digit is 6. Example 2 readings = [6] return = 6 The reading 6 is already a single digit, so the final array remains [6] and the result is 6. Example 3 readings = [3,12,23,32,0] return = 5 The readings reduce to [3,3,5,5,0]. Digits 3 and 5 each occur twice, so the tie is resolved in favor of the higher digit, 5. Constraints Every value in readings is a non-negative integer.
Reported by candidates. Source: FastPrep
Pattern and pitfall
What it really reduces to: the digital root. Repeated digit sums of a non-negative n give 0 if n is 0, otherwise 1 + (n - 1) % 9. No loop needed, and it handles huge values in O(1). Then keep a count array of size 10, and scan digits from 0 to 9 using >= when comparing counts, so ties resolve to the higher digit. The common pitfall is the zero case. A naive n % 9 returns 0 for 9, 18, and so on, which is wrong because those reduce to 9. Another slip is using strict > in the final scan and returning the lower digit on a tie, which fails Example 3. A simulation with a while loop also passes the stated limit, so don't panic if you blank on the formula. If you freeze on the live OA, StealthCoder is the hedge that gives you the clean version fast.
The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.
You can drill Most Frequent Reduced Digit 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. Built for the candidate who saw this exact problem leak two days before his OA and wondered if anyone had a play.
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Most Frequent Reduced Digit FAQ
How hard is the Airbnb Most Frequent Reduced Digit problem really?+
Easy. It's a digital root plus a frequency count. The story about sensors is noise. If you can write a digit-sum loop and a count array of size 10, you can finish it. The only trap is the tie-breaking rule and the zero case.
What's the trick to reducing a number to a single digit fast?+
Use the digital root formula. If n is 0, the result is 0. Otherwise it's 1 + (n - 1) % 9. That maps 9 to 9 and 18 to 9 correctly. You can also loop summing digits until under 10, which is fine here.
How do I handle ties between digits?+
Build a count array indexed 0 through 9. Iterate from 0 up to 9 and update your best answer whenever the current count is greater than or equal to the best count. That way the highest digit wins any tie, matching Example 3 where 3 and 5 tie and 5 is returned.
Do I need to worry about performance?+
Not really. The stated bound allows up to O(n^2), and this solution is O(n) with the formula or O(n log v) with a digit-sum loop. Either is well inside the limit. Don't spend time optimizing past a single pass over the array.
How do I prepare for this in 48 hours?+
Write the digital root both ways, loop and formula, and test on 0, 9, 18, and a single-element array. Then run the three given examples by hand. Pay attention to the tie rule. That covers every edge this problem has, and it takes under an hour.