Snapchat·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Snapchat SWE interview with a binary search problem that's basically aggressive cows in disguise. Not the hardest round I've had but you need to know your search-on-answer patterns cold.

Questions Asked (1)

Given k people and an array where each element is the number of seats in a room, place all k people across the rooms (treating seats as one long concatenated sequence) so that the minimum gap between any two adjacent people is maximized. Return that maximum possible minimum distance.

Algorithms & Data Structures

Author's notes

Took me a beat to realize this was just binary search on the answer.

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AI HintsAI Generated

Suggested Approach

This is a classic 'maximize the minimum' problem that can be solved using binary search on the answer combined with a greedy feasibility check. First, define the search space for the minimum gap (from 1 to total seats/(k-1)), then for a given gap, check if it's possible to place k people with at least that gap by greedily placing each person as early as possible. The maximum feasible gap is the answer.

Pro tip: Clarify with the interviewer whether the gap is measured as the number of seats between people or the distance between their positions (difference in indices). Also, mention that the greedy check runs in O(n) time, making the overall solution O(n log(total_seats)) which is efficient.

1. Understand the problem and define the search space

Recognize that we need to maximize the minimum distance between any two adjacent people. The answer lies between 1 and total_seats/(k-1), where total_seats is the sum of all seats.

2. Design a feasibility check function

Given a candidate minimum gap d, determine if it's possible to place k people such that each adjacent pair is at least d apart. Use a greedy approach: place the first person at position 0, then for each subsequent person, place them at the earliest position that is at least d away from the previous person.

3. Implement binary search on the answer

Perform binary search over the possible gap values. For each mid value, run the feasibility check. If feasible, search for a larger gap; otherwise, search for a smaller gap.

4. Return the maximum feasible gap

After binary search converges, return the largest gap for which the feasibility check returns true.

Key Points to Mention

Binary search on the answer (the minimum gap) to efficiently find the maximum possible value.
Greedy feasibility check: place each person as early as possible to maximize remaining space.
Time complexity: O(n log(total_seats)) where n is the number of rooms, and space complexity O(1) if we don't modify the input.
Handling edge cases: k=1 (no gap needed, return 0 or infinity depending on definition), k > total_seats (impossible, but problem likely guarantees k <= total_seats).
Clarify the definition of gap: distance between positions (difference in indices) vs. number of empty seats between.
Use prefix sums to quickly map a position in the concatenated sequence to a room and offset, enabling O(1) access during the greedy check.

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