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Meta·Software Engineer·Onsite - Coding / Algorithms·Senior

SeniorPrefer not to say
May 2026

Summary

Meta SWE interview with two fairly meaty algorithm problems back to back. The in-place merge had a nasty follow-up and the string grouping part was more subtle than it looked at first glance.

Questions Asked (4)

Q1

You have a sorted array A with some valid integers followed by sentinel (None) slots, and a separate sorted array B. You don't know how many valid elements are in A or how many sentinels there are. Implement an in-place merge into A in O(m+n) time using O(1) extra space. Cover edge cases like duplicates, negatives, and empty arrays.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

My first instinct was to scan from the front and I actually started coding that before the interviewer let me keep going for a bit and then asked me to think about what happens when you overwrite a value you haven't used yet.

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AI HintsAI Generated

Suggested Approach

First, find the number of valid elements in A by binary searching for the first sentinel (None), since A is sorted. Then, perform an in-place merge from the end of the combined array, comparing the largest valid elements from A and B and placing them in the sentinel slots, moving backwards to avoid overwriting. This achieves O(m+n) time and O(1) extra space.

Pro tip: Clarify with the interviewer whether the sentinel is a specific value (like None) and if the array has a fixed total length; this ensures you handle the sentinel detection correctly. Also, mention that if the sentinel is a large value, you can treat it as infinity and merge from the end without binary search, but binary search is needed if sentinel is None.

1. Clarify assumptions and edge cases

Confirm the representation of sentinel (e.g., None), the total length of A, and whether B is already sorted. Discuss edge cases: empty A, empty B, all sentinels, duplicates, negatives.

2. Find the number of valid elements in A

Use binary search to find the index of the first sentinel (None) in A, which gives the count of valid elements m. If no sentinel, m is the full length.

3. Merge from the end

Set pointers i = m-1 (last valid in A), j = n-1 (last in B), and k = m+n-1 (last position in A). Compare A[i] and B[j], place the larger at A[k], and decrement the corresponding pointer and k.

4. Handle remaining elements

If B still has elements after A is exhausted, copy them into the remaining positions of A. If A still has elements, they are already in place.

5. Analyze complexity and test

State that time is O(m+n) and space is O(1). Walk through examples including duplicates, negatives, and empty arrays to verify correctness.

Key Points to Mention

  • Binary search to find the first sentinel in O(log m) time, which is dominated by O(m+n) merge.
  • In-place merge from the end to avoid overwriting valid elements in A.
  • Handling duplicates by using stable comparison (e.g., >= or >) and ensuring all elements are placed.
  • Edge cases: A empty (all sentinels), B empty, both empty, all negatives, duplicates.
  • Time complexity O(m+n) and space complexity O(1).
  • Alternative if sentinel is a large value (e.g., infinity): no binary search needed, just merge from end.

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

Q2

As a follow-up to the in-place merge: can you merge by scanning forward from the beginning of both arrays instead of from the end? Why or why not, and give a concrete example showing what breaks.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This tripped me up more than it should have.

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AI HintsAI Generated

Suggested Approach

Explain that forward merging from the beginning overwrites unprocessed elements in the first array, making it impossible to merge in-place without extra space. Provide a concrete example showing the overwrite and the resulting incorrect output. Conclude that backward merging is necessary for in-place operation.

Pro tip: Mention that while forward merging fails for in-place, it works if you allocate a new array; this shows you understand the trade-off between space and time.

1. Restate the problem

Clarify that the goal is to merge two sorted arrays in-place, where the first array has enough extra space at the end to hold the second array's elements.

2. Explain forward merging

Describe the standard merge algorithm: compare the first elements of both arrays, pick the smaller, and place it at the next position in the merged array, advancing the pointer.

3. Identify the flaw

Point out that when writing to the first array from the beginning, you overwrite elements that haven't been compared yet, losing data and corrupting the merge.

4. Provide a concrete example

Use a simple example like nums1 = [1,3,5,0,0,0], m=3, nums2 = [2,4,6], n=3. Show that after placing 1 and 2, the next write would overwrite 3, leading to incorrect result.

5. Conclude and contrast

State that backward merging avoids overwriting because it fills from the end, where there is empty space, and thus is the correct in-place approach.

Key Points to Mention

  • In-place merge requires O(1) extra space.
  • Forward merging overwrites unprocessed elements in the first array.
  • Backward merging fills from the end, avoiding overwrites.
  • Concrete example: nums1 = [1,3,5,0,0,0], nums2 = [2,4,6].
  • Alternative: forward merging works if a new array is allocated (O(m+n) space).
  • Time complexity remains O(m+n) for both approaches, but space differs.

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

Q3

Given an array of lowercase strings, group together strings that are equivalent under a uniform cyclic letter shift mod 26 and have the same length. For example 'abc' and 'bcd' are in the same group, as are 'az' and 'ba'. Define a canonical key and implement the grouping.

Algorithms & Data Structures
Author's notes

Liked this problem.

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AI HintsAI Generated

Suggested Approach

First, clarify that strings are equivalent if one can be transformed into the other by shifting each character by the same amount modulo 26, and they must have the same length. Then, define a canonical key for each string by normalizing it relative to its first character (e.g., shift so the first character becomes 'a'), and use a hash map to group strings by this key. Finally, implement the solution efficiently in O(N*L) time and O(N*L) space, where N is the number of strings and L is the average length.

Pro tip: Mention that the canonical key can be computed by shifting each character by (s[0] - 'a') mod 26, but be careful with the direction of the shift to ensure consistency. Also, note that strings of different lengths cannot be equivalent, so you can skip them or handle them separately.

1. Understand the problem and clarify equivalence

Confirm that two strings are equivalent if they have the same length and there exists a uniform cyclic shift (mod 26) that transforms one into the other. Discuss edge cases like empty strings or single-character strings.

2. Define a canonical key

For each string, compute a key by shifting all characters so that the first character becomes 'a'. For example, for 'abc', shift by -0 (since 'a' is already 'a'), giving 'abc'; for 'bcd', shift by -1, giving 'abc'. This key uniquely identifies the equivalence class.

3. Group strings using a hash map

Iterate through the array, compute the canonical key for each string, and append the string to a list in a hash map keyed by the canonical key. Finally, return the lists of grouped strings.

4. Analyze complexity and optimize

Explain that the time complexity is O(N*L) where N is the number of strings and L is the maximum length, since computing each key takes O(L) time. Space complexity is O(N*L) to store the groups. Mention that this is optimal for the given constraints.

5. Test with examples and edge cases

Walk through the provided examples ('abc' and 'bcd', 'az' and 'ba') to verify the key computation. Also consider cases like strings of different lengths, empty strings, and all strings being equivalent.

Key Points to Mention

  • Equivalence relation: same length and uniform cyclic shift mod 26.
  • Canonical key: shift each string so its first character is 'a' (or any fixed character).
  • Hash map for grouping: key -> list of strings.
  • Time and space complexity: O(N*L) time, O(N*L) space.
  • Edge cases: different lengths, empty strings, single-character strings.
  • Alternative: use tuple of differences between consecutive characters as key, but first-character normalization is simpler.

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

Q4

What is the time and space complexity of your string grouping solution, and how does it handle mixed-length strings or edge cases like empty strings?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

Pretty quick exchange.

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AI HintsAI Generated

Suggested Approach

Start by clearly stating the time and space complexity of your solution, then explain how it handles mixed-length strings and edge cases like empty strings. Provide concrete examples to illustrate your reasoning and discuss any trade-offs you considered.

Pro tip: Explicitly mention that you tested edge cases such as empty strings, single-character strings, and strings with varying lengths, and explain how your solution avoids common pitfalls like incorrect grouping due to length differences.

1. State the complexity

Clearly articulate the time and space complexity of your solution, using Big O notation, and briefly justify why it is that complexity.

2. Explain handling of mixed-length strings

Describe how your algorithm groups strings of different lengths, ensuring that grouping is based on the correct criteria (e.g., sorted characters) and not inadvertently affected by length.

3. Address edge cases

Discuss how your solution handles edge cases such as empty strings, strings with only whitespace, or very long strings, and mention any special considerations.

4. Provide examples

Walk through a concrete example with mixed-length strings and an empty string to demonstrate the correctness and efficiency of your approach.

5. Discuss trade-offs

Mention any trade-offs you made (e.g., using extra space for a hash map to achieve linear time) and why they are acceptable for the given problem.

Key Points to Mention

  • Time complexity: O(N * K log K) where N is the number of strings and K is the maximum length, due to sorting each string.
  • Space complexity: O(N * K) for storing the grouped strings in a hash map.
  • Handling mixed-length strings: grouping is based on sorted characters, so length does not affect grouping correctness.
  • Edge cases: empty strings are grouped together; strings with different lengths but same characters after sorting are grouped correctly.
  • Use of a hash map with sorted string as key ensures efficient grouping.
  • Potential optimization: use character count array instead of sorting to achieve O(N * K) time.

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