J.P. Morgan·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Got a coding problem from J.P. Morgan for a software engineer role, pretty algorithmic, the kind of thing that looks manageable until you actually sit down with it.

Questions Asked (1)

You have an array of positive integers representing item quantities. Split the array at some index j (1 <= j < n) into two non-empty groups. Before splitting, you can increment or decrement any element by 1 any number of times, but every element must stay positive. Find the minimum number of such operations so that both groups have equal total quantity.

Algorithms & Data Structures

Author's notes

My first instinct was binary search on the answer, which was wrong.

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AI HintsAI Generated

Suggested Approach

First, clarify the problem and constraints, then derive a mathematical formulation. The key is to recognize that the minimum operations equal the minimum absolute difference between the sum of the first part and the sum of the second part, achievable by adjusting elements. Use prefix sums and consider parity to compute this efficiently in O(n).

Pro tip: Mention that you would validate the solution with edge cases like n=2, all elements equal, or large values, and discuss time/space complexity trade-offs. This shows attention to detail and practical engineering mindset valued at J.P. Morgan.

1. Understand the problem

Restate the problem in your own words and ask clarifying questions about constraints (e.g., array size, value ranges). Confirm that operations can be applied to any element before splitting, and that elements must remain positive.

2. Mathematical formulation

Let total sum S. For a split at index j, let L be the sum of the left part and R be the sum of the right part. The minimum operations to equalize is |L - R|, because you can transfer units between sides by incrementing/decrementing elements. Thus, minimize |L - R| over all j.

3. Algorithm design

Compute prefix sums to get L for each j, and R = S - L. The answer is min over j of |2*L - S|. This can be done in O(n) time and O(1) extra space. Note that the positivity constraint does not affect the minimum because you can always adjust elements without making them non-positive if the target difference is achievable.

4. Complexity and edge cases

Analyze time and space complexity: O(n) time, O(1) space. Discuss edge cases: n=2, all elements equal, large values causing overflow (use long), and the fact that the minimum difference might be 0 if S is even and a prefix sum equals S/2.

5. Implementation and testing

Write pseudocode or code, then walk through a small example. Suggest testing with random cases against a brute-force solution for small n to verify correctness.

Key Points to Mention

Prefix sums to compute left and right sums efficiently.
Minimizing |2*L - S| over all split points.
Time complexity O(n) and space complexity O(1).
Handling large sums with 64-bit integers.
The positivity constraint does not change the minimum operations.
Edge cases: n=2, all elements equal, total sum odd/even.

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