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Google·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Jun 2026

Summary

Two coding problems in what felt like a pretty standard Google SWE round. Both had follow-ups that pushed you further, which I wasn't fully ready for.

Questions Asked (2)

Q1

You have an array representing link weights in a gold chain. Remove exactly one link, reconnect the rest in order, then make a single cut to split the chain into two non-empty parts with equal total weight. Return true if any valid (removed index, cut position) pair exists. Follow-up: return all such valid pairs.

Algorithms & Data Structures
Author's notes

Prefix sums are the obvious tool here and I got there pretty fast.

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AI HintsAI Generated

Suggested Approach

First, clarify the problem constraints and edge cases (e.g., array size, negative weights, multiple valid pairs). Then, derive an efficient algorithm: compute total sum, and for each possible removed index, check if the remaining array can be split into two equal-sum parts using prefix sums and a hash set. For the follow-up, collect all valid pairs by iterating over all removal indices and cut positions, ensuring O(n^2) or better if possible.

Pro tip: Mention that removing a link changes the total sum, so the target for each part is (total - removed_weight)/2. Also, note that if the total minus removed weight is odd, no split is possible, which prunes many cases early.

1. Clarify and Validate

Ask about constraints: array size, weight range (negative?), whether multiple valid pairs should be returned, and if the removed link can be at any position. Confirm that after removal, the remaining links stay in order.

2. Brute Force Baseline

Describe a straightforward O(n^2) approach: for each removal index i, compute the remaining array, then for each possible cut position j, check if the sum of the left part equals the sum of the right part. This establishes correctness and helps identify optimizations.

3. Optimize with Prefix Sums

Precompute prefix sums of the original array. For a given removal index i, the remaining array's prefix sums can be derived in O(1) per cut position. Use a hash set or two-pointer technique to check for a cut where left sum equals half of (total - weight[i]).

4. Handle Follow-up: All Valid Pairs

To return all valid pairs, iterate over all removal indices i, and for each, find all cut positions j that satisfy the condition. Collect pairs (i, j) in a list. Discuss time complexity: O(n^2) worst-case, but can be optimized to O(n) per removal using prefix sums and a hash map of prefix sums to indices.

5. Test and Edge Cases

Walk through examples: empty array, single element, all zeros, negative weights, multiple valid pairs. Verify that the algorithm handles cases where no valid pair exists and returns false or empty list accordingly.

Key Points to Mention

  • Total sum and target sum calculation: after removing a link, the remaining total must be even for a split to exist.
  • Prefix sums to efficiently compute sums of subarrays after removal.
  • Time and space complexity trade-offs: O(n^2) brute force vs O(n) per removal with prefix sums and hash set.
  • Edge cases: negative weights, zeros, multiple valid pairs, and no valid pair.
  • Follow-up: returning all pairs requires careful iteration and collection, possibly using a list of results.
  • Clarifying questions: whether the removed link can be at the ends, and if the cut must be between links (not removing another link).

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

Q2

Given an integer array, find indices i and j where the values at both endpoints are equal and the subarray sum from i to j is maximized. Return one such pair. Follow-up: can you do it in O(1) extra space?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

My first instinct was a hashmap to store the first occurrence of each value, then scan and track the max sum.

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AI HintsAI Generated

Suggested Approach

Clarify the problem constraints (e.g., array size, value range, whether indices can be the same) and then propose an efficient solution. For the main solution, use a hash map to store the first occurrence of each value and track the maximum subarray sum between equal endpoints. For the follow-up, discuss how to achieve O(1) extra space by sorting or using in-place techniques, while noting trade-offs.

Pro tip: Mention that the O(1) space solution likely involves sorting the array while preserving original indices, which takes O(n log n) time but meets the space constraint. This shows you understand the trade-off between time and space.

1. Clarify the problem

Ask about constraints: array size, value range, whether i and j can be the same, and if the subarray must be non-empty. Confirm that we need to return any valid pair.

2. Brute force and optimize

Start with a brute force O(n^2) approach to ensure understanding, then optimize using a hash map to store first occurrence and prefix sums to compute subarray sums in O(1).

3. Handle edge cases

Consider cases like all elements distinct, all elements equal, negative numbers, and large arrays. Ensure the solution returns a valid pair or handles no solution gracefully.

4. Address the follow-up

For O(1) extra space, propose sorting the array with indices, then scanning to find equal values and compute sums using prefix sums stored in the sorted array. Discuss time-space trade-off.

5. Analyze complexity and trade-offs

State time and space complexity for both solutions. Compare the hash map approach (O(n) time, O(n) space) with the sorting approach (O(n log n) time, O(1) space).

Key Points to Mention

  • Use of hash map to store first occurrence of each value and prefix sums to compute subarray sums efficiently.
  • Handling negative numbers: prefix sums can decrease, so we need to track the maximum sum correctly.
  • Edge cases: empty array, single element, all distinct elements, all same elements.
  • Time and space complexity analysis: O(n) time and O(n) space for hash map; O(n log n) time and O(1) space for sorting approach.
  • Trade-offs between time and space: hash map is faster but uses extra space; sorting is slower but meets O(1) space.
  • Clarifying questions: whether indices can be the same, if the subarray must be non-empty, and if the array can be modified.

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