Use a sweep line algorithm: create events for each interval start (+1) and end+1 (-1), sort them, and track the running count to find the maximum. For the follow-up with bounded endpoints M, use a difference array of size M+2 to accumulate coverage in O(N + M) time, then scan for the max.
Pro tip: Clarify whether intervals are inclusive and whether endpoints can be large; mention that the sweep line approach handles large coordinates efficiently, while the difference array is optimal when M is small. Also, discuss tie-breaking and edge cases like overlapping intervals or single-point intervals.
Confirm that intervals are inclusive [a, b], endpoints are positive integers, and any integer with maximum coverage is acceptable. Ask about constraints on N and endpoint values.
Explain creating events: for each interval, add (a, +1) and (b+1, -1). Sort events by coordinate, then iterate while maintaining a running sum and tracking the maximum and the coordinate where it occurs.
State time complexity O(N log N) due to sorting, space O(N). Mention that this works for arbitrarily large endpoints.
If all endpoints ≤ M, use a difference array of size M+2: for each interval, diff[a] += 1 and diff[b+1] -= 1. Then compute prefix sums to get coverage at each integer and find the max. Time O(N + M), space O(M).
Handle ties by returning any max; consider intervals with a > b (invalid) or single points. For very large N but small M, difference array is better; for large M, sweep line is better.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.