Microsoft·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026

Summary

Microsoft SWE interview with a math-heavy algorithm problem that felt more like a number theory puzzle than a coding exercise. Not the LeetCode grind I was expecting.

Questions Asked (1)

A service generates consecutive integer arrays starting at some positive integer k, with length m >= 1. Given a target sum s, find how many distinct starting values k produce at least one valid array. Provide an algorithm, prove its correctness, and analyze time and space complexity. For example, s = 10 yields k = 1 and k = 10.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I knew the sum of a consecutive sequence has a closed form but blanked on how to turn that into a clean enumeration.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Model the sum of a consecutive integer array as m*(2k + m - 1)/2 = s, then derive constraints on k and m to reduce the problem to counting valid divisors or factor pairs. Iterate over possible m values up to O(sqrt(s)) and check if the corresponding k is a positive integer, ensuring each k is counted only once.

Pro tip: Emphasize that the problem reduces to counting the number of odd divisors of s, but be prepared to explain the derivation and handle edge cases like s=1 or s=0. This shows you can optimize beyond the straightforward O(sqrt(s)) approach.

1. Understand the problem and define variables

Clarify that the array is [k, k+1, ..., k+m-1] with k >= 1, m >= 1, and sum s. Write the sum formula: s = m*(2k + m - 1)/2.

2. Derive conditions for valid k and m

Rearrange to 2s = m*(2k + m - 1). Since 2k + m - 1 = 2s/m, we need m to divide 2s, and k = (2s/m - m + 1)/2 must be a positive integer. This implies m and 2s/m have opposite parity? Actually, 2k + m - 1 is an integer, so m must divide 2s, and the parity condition ensures k is integer.

3. Design an efficient algorithm

Iterate m from 1 to floor(sqrt(2s)) (or up to s) and for each divisor m of 2s, compute k. Alternatively, note that the number of valid k equals the number of odd divisors of s, and compute that in O(sqrt(s)) time.

4. Prove correctness

Show that every valid array corresponds to a unique m and k satisfying the conditions, and that the algorithm counts exactly those. Use the bijection between valid k and odd divisors of s.

5. Analyze complexity and edge cases

Time complexity O(sqrt(s)) for divisor enumeration, space O(1). Discuss edge cases: s=1 (only k=1), s=2 (only k=2), and ensure no double counting.

Key Points to Mention

Sum formula for arithmetic series: s = m*(2k + m - 1)/2
Derivation that m must divide 2s and parity condition for k to be integer
Optimization: number of valid k equals number of odd divisors of s
Time complexity O(sqrt(s)) and space O(1)
Handling edge cases like s=1, s=2, and ensuring k >= 1
Proof of correctness via bijection or direct verification

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.