TikTok·Product Manager·Technical Phone Screen·Senior

Senior

Jun 2026

Summary

TikTok PM interview with a surprisingly algorithmic problem, which felt a bit out of left field for a product role. The question was more CS-flavored than anything I'd prepped for.

Questions Asked (1)

You're given a palindromic string of lowercase letters. Replace exactly one character so the result is no longer a palindrome and is the lexicographically smallest string possible. Return the new string, or an empty string if it's impossible. Walk through your algorithm, edge cases like single-character strings, and the time and space complexity.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

Took me a moment to even accept this was the question in a PM interview.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem constraints and edge cases, then propose a greedy algorithm that scans from the left and replaces the first character that is not 'a' with 'a', unless the string consists entirely of 'a's, in which case replace the last character with 'b'. Walk through examples to validate the approach and discuss time and space complexity.

Pro tip: Emphasize that the greedy choice is optimal because changing the earliest possible character to the smallest possible letter yields the lexicographically smallest result, and always consider edge cases like single-character strings and all-'a' strings.

1. Clarify constraints and edge cases

Confirm that the input is a palindrome, consists of lowercase letters, and that exactly one character must be replaced. Discuss edge cases: length 1 (impossible), length 2 (e.g., 'aa' -> 'ab'), and strings with all 'a's.

2. Design a greedy algorithm

Scan the string from left to right. For each character, if it is not 'a', replace it with 'a' and return the result. If no such character exists (all 'a's), replace the last character with 'b' and return.

3. Validate with examples

Test the algorithm on examples like 'abba' -> 'aaba', 'aa' -> 'ab', 'aba' -> 'aaa'? Wait, 'aba' -> 'aaa' is still a palindrome, so need to check: actually 'aba' -> 'aaa' is palindrome, so greedy fails? Let's re-evaluate: For 'aba', first non-'a' is 'b' at index 1, replace with 'a' gives 'aaa' which is palindrome. So greedy needs adjustment: if replacing the first non-'a' with 'a' results in a palindrome, we might need to replace a different character. But the problem says exactly one character replacement, and result must not be palindrome. So for 'aba', replacing 'b' with 'a' gives 'aaa' (palindrome), so not allowed. Instead, we could replace the last 'a' with 'b'? That gives 'abb' which is not palindrome and lexicographically smallest? 'abb' vs 'aaa'? 'aaa' is palindrome so invalid. So we need to find the smallest string that is not palindrome. For 'aba', possible replacements: index0: 'a'->'b' gives 'bba' (not palindrome), 'a'->'c' gives 'cba', etc. index1: 'b'->'a' gives 'aaa' (palindrome), 'b'->'c' gives 'aca' (palindrome? 'aca' is palindrome), 'b'->'d' gives 'ada' (palindrome), so any replacement at index1 yields palindrome because it's the middle character. index2: 'a'->'b' gives 'abb' (not palindrome), 'a'->'c' gives 'abc' (not palindrome). So smallest is 'abb'? Compare 'abb' and 'bba': 'abb' is smaller. So for odd-length palindromes, the middle character cannot be changed to break palindrome because it's symmetric. So the greedy approach must skip the middle character if length is odd. So algorithm: scan from left to right, but if index is the middle index (when length is odd), skip it. Replace the first non-'a' character that is not the middle with 'a'. If all such characters are 'a', then replace the last character with 'b'. This works for 'aba': middle index 1 is 'b', skip; then index0 is 'a', index2 is 'a', so all non-middle are 'a', so replace last with 'b' -> 'abb'. Correct. So step 3: Validate with examples like 'abba' -> 'aaba', 'aba' -> 'abb', 'aa' -> 'ab', 'a' -> ''.

4. Analyze complexity

The algorithm scans the string once, so time complexity is O(n). Space complexity is O(n) if we create a new string, or O(1) if we modify in place (but strings are immutable in many languages, so O(n) space for the result).

5. Summarize and conclude

Restate the algorithm, confirm it handles all edge cases, and mention that if the string length is 1, return an empty string as it's impossible.

Key Points to Mention

Greedy strategy: replace the earliest possible character with 'a' to minimize lexicographically, but skip the middle character for odd-length palindromes.
Edge cases: single-character string (impossible, return empty string), all 'a's (replace last with 'b'), and odd-length palindromes where the middle character cannot be changed to break palindrome.
Time complexity O(n) and space complexity O(n) due to string immutability.
Proof of correctness: changing the earliest character to the smallest possible letter yields the lexicographically smallest string, and skipping the middle ensures the result is not a palindrome.
Handling of exactly one replacement: ensure that the replacement actually changes the character (e.g., if the character is already 'a', we don't replace it with 'a' because that wouldn't change the string).
Consideration of product management aspects: clarifying requirements, discussing trade-offs between simplicity and edge case handling, and communicating the solution clearly.

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