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Meta·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jun 2026

Summary

Meta SWE coding round with two questions, both algorithmic. Nothing behavioral, just pure problem solving under the clock. Felt like a pretty standard technical screen but the second question had some tricky edge cases worth thinking through.

Questions Asked (2)

Q1

Given a string, can you determine whether removing at most one character makes it a palindrome?

Algorithms & Data Structures
Author's notes

Two-pointer from both ends, and when you hit a mismatch you try skipping one character from either side and check if the remainder is a palindrome.

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AI HintsAI Generated

Suggested Approach

Use a two-pointer technique to compare characters from both ends, and when a mismatch occurs, check if skipping either the left or right character results in a palindrome. This yields an O(n) time and O(1) space solution.

Pro tip: Clarify that 'at most one' includes zero removals, and mention that the two-pointer approach is optimal for this problem, avoiding unnecessary string copying.

1. Clarify the problem

Confirm that removing at most one character means zero or one removal is allowed, and that the string can contain any characters.

2. Initialize two pointers

Set left pointer at the start and right pointer at the end of the string.

3. Compare and handle mismatch

While left < right, if characters match, move both pointers inward. If they don't match, check if the substring skipping left or skipping right is a palindrome.

4. Check palindrome validity

Write a helper function that checks if a substring is a palindrome using two pointers, and use it on the two possible substrings after a mismatch.

5. Return result

If either check returns true, the answer is true; otherwise, false. If no mismatch occurs, return true.

Key Points to Mention

  • Time complexity: O(n) because each character is visited at most twice.
  • Space complexity: O(1) as no extra data structures are used.
  • Handling edge cases: empty string, single character, already palindrome, and strings requiring removal at different positions.
  • The two-pointer technique efficiently narrows down the problem.
  • The helper function for palindrome check can be reused.
  • Avoid modifying the string or creating substrings to keep space complexity low.

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

Q2

Given a binary tree, return its vertical order traversal, where nodes in the same column are listed top to bottom, and ties in row and column follow left-to-right BFS order.

Algorithms & Data Structures
Author's notes

BFS with a column tracker is the right move.

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AI HintsAI Generated

Suggested Approach

Use BFS to traverse the tree level by level, tracking each node's column index. Store nodes in a hash map keyed by column, then sort the columns and output the nodes in each column in the order they were added.

Pro tip: Mention that BFS naturally handles the top-to-bottom and left-to-right ordering, and that using a hash map with sorted keys ensures columns are processed in the correct order.

1. Clarify the problem

Confirm that vertical order means grouping nodes by column index, with nodes in the same column ordered top-to-bottom, and ties broken by left-to-right BFS order.

2. Choose BFS with column tracking

Use a queue for BFS, storing each node along with its column index. Start with the root at column 0.

3. Collect nodes by column

Use a hash map to map column indices to lists of node values. For each node dequeued, append its value to the list for its column.

4. Sort and output columns

After BFS, sort the column keys in ascending order and concatenate the lists to form the final result.

5. Analyze complexity

State that time complexity is O(n log n) due to sorting columns (or O(n) if using a tree map), and space complexity is O(n) for the queue and map.

Key Points to Mention

  • BFS ensures top-to-bottom and left-to-right ordering within the same column.
  • Use a hash map (or dictionary) to group nodes by column index.
  • Column indices can be negative; handle sorting accordingly.
  • Time complexity: O(n log n) due to sorting columns; can be O(n) with a balanced BST.
  • Space complexity: O(n) for the queue and map.
  • Edge cases: empty tree, skewed tree, nodes with same column but different rows.

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