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Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
May 2026

Summary

Meta SWE coding screen, one question on string manipulation. Pretty standard stuff but the edge cases are where it gets you.

Questions Asked (1)

Q1

Given a string, can you determine whether it becomes a palindrome if you remove at most one character? Only alphanumeric characters count and case doesn't matter.

Algorithms & Data Structures
Author's notes

Two pointer approach is pretty natural here but the tricky part is what to do when you hit a mismatch.

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AI HintsAI Generated

Suggested Approach

Use a two-pointer technique starting from both ends, moving inward while characters match. When a mismatch occurs, check if skipping either the left or right character results in a palindrome, allowing at most one removal. Preprocess the string by filtering out non-alphanumeric characters and converting to lowercase.

Pro tip: Clarify edge cases upfront, such as empty strings or strings with only one character, and mention that the solution runs in O(n) time with O(1) extra space (if preprocessing is done in-place or with O(n) space if creating a new string). This shows attention to detail and efficiency.

1. Clarify and preprocess

Confirm that only alphanumeric characters are considered and case is ignored. Preprocess the string by removing non-alphanumeric characters and converting to lowercase.

2. Initialize two pointers

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

3. Compare and move pointers

While left < right, if characters match, move both pointers inward. If they don't match, proceed to check for palindrome by skipping one character.

4. Check skip left or right

When a mismatch occurs, check if the substring from left+1 to right is a palindrome OR the substring from left to right-1 is a palindrome. If either is true, return true; otherwise, return false.

5. Return result

If the loop completes without mismatches, return true. If a mismatch was found and neither skip works, return false.

Key Points to Mention

  • Two-pointer technique for palindrome checking
  • Handling at most one removal by trying both skips
  • Time complexity: O(n) where n is the length of the string
  • Space complexity: O(1) if preprocessing is done in-place, otherwise O(n) for the cleaned string
  • Edge cases: empty string, single character, strings with only non-alphanumeric characters
  • Case insensitivity and alphanumeric filtering

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