Bloomberg·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

Bloomberg coding screen for a software engineer role, pretty focused on string manipulation and complexity analysis. The follow-up about outputting an actual index to delete was the part that separated the question from a standard palindrome check.

Questions Asked (1)

Given a string, can you determine whether it becomes a palindrome after removing at most one character? Return true or false, state the time and space complexity, and as a follow-up: if a deletion is possible, return one valid index to delete.

Algorithms & Data Structures

Author's notes

The base case is pretty textbook, two pointers moving inward until you find a mismatch, then you try skipping one side or the other and see if the remainder is a palindrome.

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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 results in a palindrome. If either does, return true (and the index to delete); otherwise, return false. This yields O(n) time and O(1) space.

Pro tip: After finding a mismatch, verify both deletion options with a helper function that checks if a substring is a palindrome. For the follow-up, return the index of the character that, when deleted, makes the string a palindrome; if both work, either is acceptable, but mention that you can return the first one found.

1. Clarify and confirm

Restate the problem: determine if a string can become a palindrome by deleting at most one character. Confirm that you need to return a boolean and, for the follow-up, an index to delete if possible.

2. Two-pointer scan

Initialize left and right pointers at the start and end of the string. Move them inward while characters match.

3. Handle mismatch

When a mismatch occurs, check if the substring skipping the left character is a palindrome, or if skipping the right character is a palindrome. If either is true, return true (and the corresponding index); otherwise, return false.

4. Palindrome check helper

Implement a helper function that checks if a substring (given left and right indices) is a palindrome using two pointers, without creating new strings to maintain O(1) space.

5. Analyze complexity

State that the time complexity is O(n) because each character is visited at most a constant number of times, and space complexity is O(1) as only pointers and indices are used.

Key Points to Mention

Two-pointer technique for efficient palindrome checking.
Handling the mismatch by trying both deletion options.
Using a helper function to check palindrome on substrings without extra space.
Time complexity: O(n) due to linear scan and constant-time checks.
Space complexity: O(1) because no additional data structures are used.
For the follow-up, returning the index to delete (e.g., left or right pointer index) when a valid deletion is found.

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