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This question evaluates a candidate's competency in combinatorial counting, digit-sum properties, and algorithmic optimization for range queries within the Coding & Algorithms domain.

  • Medium
  • Roblox
  • Coding & Algorithms
  • Software Engineer

Find largest digit-sum bucket size

Company: Roblox

Role: Software Engineer

Category: Coding & Algorithms

Difficulty: Medium

Interview Round: Technical Screen

You are given two integers low and high (inclusive). Define s(x) as the sum of the decimal digits of x. For every integer x in [low, high], place x into bucket s(x). Return the size of the largest bucket (i.e., the maximum count of numbers that share the same digit sum). If multiple buckets tie, return that common size. Describe your algorithm and analyze its time and space complexity. Follow-ups: ( 1) Can you avoid iterating over every x when high − low is very large? ( 2) How would your approach change if numbers were represented in an arbitrary base b (2 ≤ b ≤ 36)?

Quick Answer: This question evaluates a candidate's competency in combinatorial counting, digit-sum properties, and algorithmic optimization for range queries within the Coding & Algorithms domain.

You are given two integers `low` and `high` (inclusive). Define `s(x)` as the sum of the decimal digits of `x`. For every integer `x` in the range `[low, high]`, place `x` into the bucket labeled `s(x)`. Return the size of the largest bucket — that is, the maximum number of integers in `[low, high]` that share the same digit sum. If several buckets tie for the maximum, return that common size. Example: for `low = 1, high = 10`, the digit sums are 1,2,3,4,5,6,7,8,9,1. The bucket for digit sum 1 contains {1, 10} (size 2); every other bucket has size 1. The answer is 2. Follow-ups to discuss (not required by the tests): (1) How could you avoid iterating over every `x` when `high - low` is astronomically large — e.g. by counting, via digit-DP, how many numbers in `[low, high]` have each possible digit sum? (2) How would the approach change for an arbitrary base `b` (2 <= b <= 36)?

Constraints

  • 1 <= low <= high <= 10^5 (iteration over the range is feasible at this scale)
  • s(x) is the sum of the base-10 digits of x
  • The answer is the maximum bucket size; ties resolve to the same common size

Examples

Input: (1, 10)

Expected Output: 2

Explanation: Digit sums 1..9 then 1 again for 10; bucket for sum 1 = {1,10} has size 2, the max.

Input: (5, 15)

Expected Output: 2

Explanation: Sum 6 = {6, 15} and sum 5 = {5, 14} each have size 2; max bucket size is 2.

Input: (19, 28)

Expected Output: 2

Explanation: Sums: 19->10, 20->2, 21->3, 22->4, 23->5, 24->6, 25->7, 26->8, 27->9, 28->10; bucket for sum 10 = {19,28} has size 2.

Input: (7, 7)

Expected Output: 1

Explanation: Single element; its bucket (sum 7) has size 1.

Input: (1, 100)

Expected Output: 10

Explanation: The most populated digit-sum bucket over 1..100 holds 10 numbers.

Input: (100, 110)

Expected Output: 2

Explanation: Crossing a hundreds boundary; the largest shared-digit-sum bucket has size 2 (e.g. sum 1 = {100, 110}).

Hints

  1. Compute the digit sum of each number by repeatedly taking x % 10 and x //= 10.
  2. Use a hash map keyed by digit sum to count how many numbers fall in each bucket, tracking the running maximum.
  3. For the large-range follow-up, think about counting numbers with a given digit sum using digit DP instead of enumerating every x: count(high, target) - count(low - 1, target).
Last updated: Jun 26, 2026

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