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Bloomberg·Software Engineer·Onsite - Coding / Algorithms·Intermediate

IntermediatePrefer not to say
Jul 2026

Summary

Bloomberg SWE coding round, two questions back to back. Nothing too wild but the linked list one took me longer than I'd like to admit.

Questions Asked (2)

Q1

Given two strings, determine whether they are anagrams of each other. Must run in O(n) time and use O(1) extra space.

Algorithms & Data Structures
Author's notes

Pretty standard stuff.

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AI HintsAI Generated

Suggested Approach

Clarify the character set (e.g., ASCII vs Unicode) and constraints, then propose a counting approach using a fixed-size array (e.g., 256 for ASCII) to tally character frequencies. Iterate through both strings, incrementing for the first and decrementing for the second, ensuring all counts return to zero. This achieves O(n) time and O(1) extra space (since the array size is constant).

Pro tip: Mention that if the character set is not fixed, O(1) space is impossible for arbitrary Unicode; showing awareness of this constraint demonstrates depth. Also, handle edge cases like different lengths early to avoid unnecessary work.

1. Clarify constraints and assumptions

Ask about the character set (e.g., ASCII, Unicode) and whether the strings can contain any characters. Confirm that O(1) extra space means constant space independent of input size.

2. Check basic edge cases

If the strings have different lengths, they cannot be anagrams; return false immediately. Also handle empty strings (both empty are anagrams).

3. Choose a counting strategy

Use a fixed-size frequency array (e.g., size 256 for ASCII) to count character occurrences. This array size is constant, satisfying O(1) space.

4. Iterate and update counts

Traverse the first string, incrementing the count for each character. Traverse the second string, decrementing the count. If any count goes negative, return false early.

5. Verify and conclude

After both traversals, check that all counts are zero. If so, return true; otherwise, false. Discuss time complexity O(n) and space O(1).

Key Points to Mention

  • Time complexity: O(n) where n is the length of the strings (assuming equal length).
  • Space complexity: O(1) because the frequency array size is fixed (e.g., 256 for ASCII).
  • Character set assumption: The solution works for a fixed character set like ASCII; for Unicode, O(1) space is not feasible.
  • Early termination: If lengths differ or a count goes negative, return false immediately.
  • Alternative approaches: Sorting (O(n log n)) or hash map (O(n) space) are less optimal; explain why counting array is better.
  • Edge cases: Empty strings, strings with repeated characters, and strings with different lengths.

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

Q2

Reorder a singly linked list so all nodes at odd indices come first, followed by all nodes at even indices, preserving relative order within each group. In-place only, O(n) time and O(1) space.

Algorithms & Data Structures
Author's notes

I stared at this for a bit before it clicked.

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AI HintsAI Generated

Suggested Approach

Use two pointers to build two separate chains for odd and even indexed nodes, then link the end of the odd chain to the head of the even chain. Maintain pointers to the tails of both chains to append nodes in O(1) time per node, ensuring O(n) time and O(1) space.

Pro tip: Clarify whether indices are 1-based or 0-based; typically odd/even refers to 1-based positions. Also, handle edge cases like empty list, single node, or two nodes explicitly to avoid null pointer exceptions.

1. Clarify and handle edge cases

Confirm the definition of odd/even indices (1-based vs 0-based) and check for edge cases: empty list, single node, or two nodes. If the list has fewer than 3 nodes, no reordering is needed.

2. Initialize pointers

Set up pointers for the odd and even chains: oddHead, oddTail, evenHead, evenTail. Initially, oddHead = head, evenHead = head.next, and set tails accordingly.

3. Traverse and build chains

Iterate through the list, alternating between odd and even nodes. For each node, append it to the corresponding chain by updating the tail's next pointer, then advance the tail.

4. Combine chains

After traversal, link the odd chain's tail to the even chain's head. Ensure the even chain's tail's next is set to null to terminate the list.

5. Return and verify

Return the head of the odd chain (which is the original head). Walk through the list to verify the order and that no cycles exist.

Key Points to Mention

  • In-place manipulation using pointer rewiring, no extra data structures.
  • Time complexity O(n) because each node is visited once.
  • Space complexity O(1) due to constant number of pointers.
  • Preservation of relative order within odd and even groups.
  • Handling of edge cases such as empty list, single node, and two nodes.
  • Potential pitfalls: null pointer dereferences, losing references, and creating cycles.

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