My first instinct was brute force: try removing each index, recompute prefix sums, check for a pivot.
First, clarify the problem constraints and edge cases, then propose an efficient solution using prefix sums to track left and right sums. Explain how to handle the removal of exactly one element by checking if removing an element can make the remaining array have a pivot index, possibly using a modified prefix sum approach.
Pro tip: Mention that you would test with arrays containing duplicates and negative numbers, as they can break naive assumptions about monotonicity. Also, discuss the time and space complexity trade-offs between a brute-force O(n^2) approach and an optimized O(n) solution.
Ask clarifying questions: Does the pivot index refer to the original array after removal? Can the pivot be at the first or last position? What if multiple removals yield a valid pivot? Confirm that exactly one element must be removed.
For each element, remove it and check if the remaining array has a pivot index by computing left and right sums for each possible pivot. This is O(n^2) time and O(1) extra space, but may be too slow for large inputs.
Compute total sum. For each index i, consider removing element i. The remaining array has sum total - arr[i]. Then check if there exists a pivot index j in the remaining array such that left sum equals right sum. Use prefix sums to check in O(1) per candidate.
Consider arrays of length 1 or 2, all negative numbers, duplicates, and cases where the pivot is at the boundary. Ensure the removal index is not the same as the pivot index in the remaining array.
State time complexity O(n) and space O(n) for prefix sums, or O(1) if using running sums. Walk through a few examples to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.