Meta·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Meta coding round focused on string manipulation with a real algorithmic twist. The palindrome question sounds easy until you actually have to return the index too, not just a boolean.

Questions Asked (1)

Given a string, can it become a palindrome by removing at most one character? Return both a true/false result and a valid index to delete. Walk through an O(n) solution and explain how you'd handle Unicode or case-insensitive comparisons.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The two-pointer part came naturally but I completely forgot they also wanted the index back until I was halfway through explaining.

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AI HintsAI Generated

Suggested Approach

Use a two-pointer approach from both ends, and when a mismatch occurs, check if skipping either the left or right character yields a palindrome. Return true and the index to delete if so; otherwise return false. For Unicode or case-insensitive comparisons, normalize the string first (e.g., using Unicode normalization and case folding) and then apply the same algorithm.

Pro tip: Mention that you'd clarify the definition of 'character' (code unit vs. grapheme) and whether the index should refer to the original or normalized string, as this affects correctness and is a common pitfall in real-world Unicode handling.

1. Clarify requirements and edge cases

Ask about input constraints, definition of character (Unicode code point vs. grapheme), case sensitivity, and whether the index should be in the original or normalized string. Discuss empty string, single character, and already palindrome cases.

2. Normalize the string if needed

If case-insensitive or Unicode normalization is required, apply appropriate transformations (e.g., toLowerCase, NFC/NFD) and keep a mapping to original indices if necessary.

3. Implement two-pointer check with skip

Use two pointers (left, right) moving inward. On mismatch, check if the substring skipping left or skipping right is a palindrome. If either is, return true and the index to delete; otherwise return false.

4. Return the index to delete

If skipping left yields a palindrome, return the left index; if skipping right yields a palindrome, return the right index. If no deletion needed, return true and -1 (or null) to indicate no deletion.

5. Analyze complexity and trade-offs

Explain that the algorithm runs in O(n) time and O(1) extra space (ignoring normalization). Discuss trade-offs of normalizing (extra space/time) and handling Unicode correctly.

Key Points to Mention

Two-pointer technique with a helper function to check palindrome in a range.
Time complexity O(n) and space complexity O(1) for the core algorithm.
Handling of Unicode: normalization forms (NFC, NFD), case folding, and grapheme clusters.
Edge cases: empty string, single character, already palindrome, multiple mismatches.
Returning the index: clarify if it's the index in the original string or after normalization.
Trade-offs: normalizing may change indices and require mapping; in-place comparison may be faster but less correct for Unicode.

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