The problem wraps an interval scheduling question in a bunch of marketplace flavor text and I spent way too long parsing what 'cohesive' actually meant.
Reframe the problem as finding the maximum subset of merchants that share a common overlapping point, then subtract from n to get the minimum removals. Sort the intervals and use a sweep line or greedy approach to find the point covered by the most intervals, which corresponds to the largest valid subset.
Pro tip: Clarify that the 'zone' is an interval [start, end] and that overlap includes touching endpoints; this avoids off-by-one errors and shows attention to detail. Also, mention that if multiple points have the same maximum coverage, any works, but the minimum removals is n minus that maximum.
Confirm that zones are intervals and that overlap means sharing at least one point. Restate the goal: remove the fewest merchants so that the remaining set has a common intersection point.
Observe that the remaining merchants must all contain some point p. Thus, the problem reduces to finding the point p covered by the maximum number of intervals. The answer is n minus that maximum.
Sort all start and end points. Use a sweep line: increment count at start, decrement after end (careful with inclusive endpoints). Track the maximum count. Alternatively, sort intervals by start and use a min-heap of end points to find the maximum overlap.
After finding the maximum number of overlapping intervals, return n minus that maximum as the minimum number of merchants to remove.
State that the sweep line approach runs in O(n log n) time due to sorting, with O(n) space for the events or heap. This is optimal for comparison-based sorting.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.