Google·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026Remote

Summary

Phone screen coding round at Google for a SWE role. The base problem was easy but it's basically just the warm-up; the real graded part is a parallel search follow-up that catches a lot of people off guard if they haven't practiced it.

Questions Asked (2)

Implement a binary search to find the first bad version given a monotonic isBadVersion(v) oracle over versions 1 to n.

Algorithms & Data Structures

Author's notes

Straightforward binary search, nothing tricky.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Use binary search to find the first version where isBadVersion returns true. Maintain a search range [1, n] and narrow it down by checking the middle version, ensuring the first bad version is always within the range. Return the left boundary when the range converges.

Pro tip: Mention that to avoid integer overflow when computing mid, use left + (right - left) / 2 instead of (left + right) / 2. Also, clarify that the problem guarantees at least one bad version, so no need to handle the case where none are bad.

1. Clarify the problem and constraints

Confirm that versions are numbered 1 to n, isBadVersion is monotonic (all versions after the first bad are bad), and there is at least one bad version. Ask if n can be large (e.g., up to 2^31-1) to discuss overflow.

2. Define the search space and invariant

Set left = 1 and right = n. The invariant is that the first bad version lies within [left, right]. Initially, this holds because at least one bad version exists.

3. Implement binary search loop

While left < right, compute mid = left + (right - left) / 2. If isBadVersion(mid) is true, the first bad version is at mid or to the left, so set right = mid. Otherwise, set left = mid + 1.

4. Return the result

When left == right, that index is the first bad version. Return left.

5. Analyze complexity and edge cases

Time complexity is O(log n) due to halving the search space. Space is O(1). Test edge cases: n=1, first bad is 1, first bad is n, and large n to check overflow handling.

Key Points to Mention

Binary search reduces time complexity from O(n) to O(log n).
Use mid = left + (right - left) / 2 to prevent integer overflow.
The monotonic property of isBadVersion ensures binary search is applicable.
The loop terminates when left == right, which is the first bad version.
Handle edge cases such as n=1 and when the first bad version is at the boundaries.
Space complexity is O(1) as only a few variables are used.

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

Now parallelize that search: partition the version range into buckets, probe the bucket boundaries concurrently to find which bucket contains the good-to-bad transition, then binary search within that bucket. Write working code.

Algorithms & Data StructuresSystem DesignTechnical Trade-offs

Author's notes

This is where I nearly ran out of time.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

First, clarify the problem: we have a version range [low, high] where low is good and high is bad, and we need to find the first bad version. The parallel approach partitions the range into k buckets, probes the boundaries concurrently to identify the bucket containing the transition, then performs a binary search within that bucket. Implement a function that takes a predicate isBad(version) and returns the first bad version, using concurrency primitives like goroutines and channels (or threads/futures) to parallelize the boundary probes.

Pro tip: Mention that the optimal number of buckets is sqrt(N) to balance the cost of parallel probing and subsequent binary search, and discuss how to handle cases where multiple boundaries are bad or good to narrow down efficiently.

1. Clarify the problem and constraints

Confirm that the version range is sorted (all good before all bad), the predicate is monotonic, and discuss the cost model: parallel probes are cheap but not free, and the number of buckets affects performance.

2. Design the parallel bucketing strategy

Divide the range [low, high] into k equal-sized buckets. Compute boundary versions (e.g., low + i*(high-low)/k for i=1..k-1) and probe them concurrently using threads/goroutines.

3. Identify the transition bucket

Collect the results of the boundary probes. Find the first boundary that is bad; the transition lies between the previous boundary (good) and this boundary (bad). If no boundary is bad, the transition is in the last bucket.

4. Binary search within the bucket

Perform a standard binary search on the sub-range [prev_good+1, first_bad] to find the exact first bad version. This can be done sequentially or in parallel if the bucket is large.

5. Implement and test the code

Write clean, working code with proper concurrency handling (e.g., sync.WaitGroup, channels). Test with edge cases: all good, all bad, transition at boundaries, and small ranges.

Key Points to Mention

Monotonicity of the predicate: all versions before the first bad are good, all after are bad.
Choice of k: sqrt(N) minimizes total probes (k parallel probes + log(N/k) binary search steps).
Concurrency primitives: use goroutines and channels (Go) or threads/futures (Java/Python) to probe boundaries in parallel.
Handling edge cases: when the first or last bucket contains the transition, and when the range size is smaller than k.
Time complexity: O(k + log(N/k)) with k parallel probes, assuming unlimited parallelism; space complexity O(k) for storing results.
Trade-offs: more buckets reduce binary search steps but increase parallel overhead; consider limited parallelism if resources are constrained.

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