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Google·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Two coding problems for a Google SWE round, nothing too wild but the second one had a subtle edge case that tripped me up a bit. Pretty standard algorithmic fare overall.

Questions Asked (2)

Q1

Given an integer, determine whether it reads the same forwards and backwards. Negative numbers should return false. Try to solve it without converting to a string.

Algorithms & Data Structures
Author's notes

The no-string constraint is the whole point of this problem.

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AI HintsAI Generated

Suggested Approach

Clarify that negative numbers return false immediately, then reverse the integer mathematically using modulo and division while checking for overflow. Compare the reversed number with the original, or reverse only half the digits to avoid overflow and optimize.

Pro tip: Mention that reversing only half the digits avoids overflow and is more efficient; also discuss handling trailing zeros and edge cases like 0 and multiples of 10.

1. Clarify constraints and edge cases

Confirm that negative numbers return false, and consider edge cases like 0, single-digit numbers, and numbers ending in 0 (e.g., 10).

2. Choose reversal strategy

Decide between reversing the entire number or just half. Reversing half avoids overflow and is more efficient.

3. Implement mathematical reversal

Use a loop with modulo 10 to extract digits and build the reversed number, or build half-reversed and compare with the remaining half.

4. Handle termination and comparison

Stop when the reversed half is greater than or equal to the remaining half, then compare appropriately (equal for even digits, or equal after dividing by 10 for odd digits).

5. Test with examples

Walk through examples like 121 (true), -121 (false), 10 (false), and 0 (true) to verify correctness.

Key Points to Mention

  • Negative numbers return false immediately.
  • Numbers ending in 0 (except 0 itself) cannot be palindromes.
  • Reversing only half the digits avoids integer overflow.
  • Time complexity is O(log10(n)) and space complexity is O(1).
  • Edge cases: 0, single-digit numbers, and maximum/minimum integer values.
  • Comparison logic for odd vs. even number of digits.

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

Q2

Given a list of intervals in any order, merge all overlapping ones and return the result sorted by start value.

Algorithms & Data Structures
Author's notes

Sort first, then walk through and merge.

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AI HintsAI Generated

Suggested Approach

Start by clarifying assumptions (e.g., intervals are closed, input may be empty). Sort intervals by start value, then iterate through them, merging overlapping intervals by comparing the current interval's start with the last merged interval's end. Return the merged list, which is naturally sorted by start.

Pro tip: After presenting the solution, mention that you can optimize to O(n log n) time and O(n) space, and discuss edge cases like empty input, single interval, and intervals that touch (e.g., [1,2] and [2,3]). This shows thoroughness and attention to detail.

1. Clarify and Validate

Ask clarifying questions about interval inclusivity, input size, and expected output format. Confirm that intervals are given as pairs of integers and that the result should be sorted by start.

2. Sort Intervals

Sort the list of intervals by their start value. This ensures that any overlapping intervals are adjacent, simplifying the merging process.

3. Merge Overlapping Intervals

Initialize an empty result list. Iterate through sorted intervals; if the result is empty or the current interval does not overlap with the last interval in result, append it. Otherwise, merge by updating the end of the last interval to the maximum of both ends.

4. Return and Analyze

Return the merged list. Discuss time complexity (O(n log n) due to sorting) and space complexity (O(n) for the output). Mention potential optimizations if the input is already sorted.

Key Points to Mention

  • Sorting by start value is crucial for O(n log n) efficiency.
  • Overlap condition: next.start <= current.end (for closed intervals).
  • Merging by updating the end to max(current.end, next.end).
  • Edge cases: empty input, single interval, intervals that just touch.
  • Time complexity: O(n log n) due to sorting; space complexity: O(n) for output.
  • Alternative approaches (e.g., using a stack) and their trade-offs.

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