First thing I did was ask about equal elements because that felt like a trap waiting to happen.
Clarify the problem constraints and edge cases, then propose a single-pass solution that maintains two counters for elements greater than and less than the pivot. Emphasize that this achieves O(n) time and O(1) space, and finally compare the counters to return the appropriate symbol.
Pro tip: Mention that you can early-exit if one counter exceeds the maximum possible count for the other, but note that this doesn't change worst-case complexity. Also, explicitly state that elements equal to the pivot are ignored, showing attention to detail.
Ask about input size, data types, and whether the array can be empty or contain duplicates. Confirm that equal elements are ignored and that the pivot is not necessarily in the array.
Explain that you will iterate through the array once, incrementing a greater counter when an element > pivot and a lesser counter when an element < pivot. Use only two integer variables for space.
Demonstrate with a simple array (e.g., [1, 5, 3, 7], pivot=4) to show how the counters update and how the final comparison yields '='.
State that time is O(n) and space is O(1). Optionally mention early termination if one count exceeds the other's maximum possible value, but clarify it doesn't improve worst-case.
Implement the solution in a chosen language, handling edge cases like empty array. Test with cases where greater > lesser, lesser > greater, and equal.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.