Walmart Labs·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Walmart Labs SWE interview, one algorithmic question that sounds easy until you actually think about the edge cases. They wanted the full treatment: algorithm, correctness argument, complexity analysis.

Questions Asked (1)

Given N bus routes each with a start time (inclusive) and end time (exclusive), find the maximum number of routes that are running at the same time. Walk through your algorithm, explain why it's correct, and give the time and space complexity.

Algorithms & Data Structures

Author's notes

My first instinct was brute force, check every pair of intervals, which I knew was wrong but I said it anyway to buy time.

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AI HintsAI Generated

Suggested Approach

Clarify the problem and edge cases, then propose an event-based sweep line algorithm: create events for each route's start (+1) and end (-1), sort them, and sweep to track the maximum concurrent routes. Explain the algorithm step-by-step, prove its correctness, and state the time and space complexity.

Pro tip: Mention that sorting end events before start events at the same timestamp correctly handles the half-open interval [start, end), and note that if times are bounded integers, a counting sort can achieve O(N + K) time.

1. Clarify the problem and constraints

Confirm the interval semantics (inclusive start, exclusive end), input format, and any constraints on N or time values. Discuss edge cases like empty input, single route, or all routes overlapping.

2. Propose the sweep line algorithm

Create events: for each route, add (start, +1) and (end, -1). Sort events by time, with end events before start events at the same time to respect the half-open interval. Sweep through events, maintaining a running count and updating the maximum.

3. Walk through an example

Illustrate with a small example (e.g., routes [1,4), [2,5), [3,6)) to show how the count changes and the maximum is found. This demonstrates understanding and catches off-by-one errors.

4. Prove correctness

Argue that the running count equals the number of active routes at any time, and since we check the count after every event, the maximum is captured. The tie-breaking rule ensures intervals ending at time t are not counted as active at t.

5. State time and space complexity

Sorting 2N events takes O(N log N) time; sweeping takes O(N) time. Space is O(N) for events. Mention that if times are small integers, counting sort can reduce time to O(N + K).

Key Points to Mention

Event-based sweep line approach with +1 for start and -1 for end
Sorting events by time, with end events before start events at the same timestamp
Half-open interval semantics: [start, end) means end time is exclusive
Time complexity: O(N log N) due to sorting; space complexity: O(N)
Alternative O(N + K) approach using counting sort if time range K is small
Edge cases: empty input, all routes overlapping, routes with same start/end times

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