JP Morgan Chase·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

JP Morgan Chase SWE interview with a classic array problem. Nothing too wild but the question has a few edge cases that can trip you up if you're not careful.

Questions Asked (1)

Given an array of distinct integers, find all pairs with the minimum absolute difference.

Algorithms & Data Structures

Author's notes

I went straight for sorting the array first, which is the right move, but I fumbled explaining why for a second.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and edge cases, then propose sorting the array to bring close values together. After sorting, a single pass can compute adjacent differences and collect all pairs achieving the minimum difference.

Pro tip: Mention that sorting is O(n log n) and is optimal for comparison-based approaches; if the interviewer asks for better, note that O(n) is possible with hashing when the value range is small, but sorting is generally preferred for its simplicity and robustness.

1. Clarify and confirm

Restate the problem to ensure understanding: distinct integers, find all pairs with minimum absolute difference. Ask about input size, value range, and whether the output order matters.

2. Sort the array

Sort the array in ascending order. This ensures that the minimum absolute difference will be between adjacent elements, reducing the problem to checking consecutive pairs.

3. Single pass to find min diff and pairs

Initialize min_diff to infinity and an empty result list. Iterate through adjacent pairs, compute the difference, and update min_diff and result accordingly: if diff < min_diff, reset result; if diff == min_diff, append the pair.

4. Return result

After the pass, return the list of pairs. Discuss time and space complexity: O(n log n) time due to sorting, O(n) space for the result in the worst case.

Key Points to Mention

Sorting brings potential minimum-difference pairs adjacent, simplifying the search.
Time complexity: O(n log n) for sorting plus O(n) for the pass, total O(n log n).
Space complexity: O(n) for the output list, O(1) extra if we ignore output.
Edge cases: array size less than 2 (return empty), all elements equally spaced (all adjacent pairs have same diff).
Alternative approach: using a hash set to track seen numbers and compute differences on the fly, but sorting is more straightforward.
The algorithm handles distinct integers; if duplicates were allowed, minimum difference could be 0 and pairs would include duplicates.

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