Maximum Concurrent Processes

You are given n process logs. Each log is an inclusive integer interval [startTime, endTime], meaning that the process runs at every minute from startTime through endTime.
Return the maximum number of processes that are running simultaneously at any integer minute.

Function
maximumConcurrentProcesses(processLogs: int[][]) → int

Examples
Example 1
processLogs = [[1,3],[2,4],[3,6]]
return = 3
At minute 3, all three inclusive intervals contain that minute, so three processes are running simultaneously.
Example 2
processLogs = [[2,6],[1,2],[3,5]]
return = 2
At minute 2, the first two processes overlap. At minutes 3 through 5, the first and third processes overlap. No minute belongs to all three intervals.

Constraints
Each entry in processLogs contains exactly two integers: [startTime, endTime].
For every process log, startTime <= endTime.
Both endpoints are inclusive.

The trick is turning each interval into two events: +1 at startTime and -1 at endTime + 1. The +1 offset matters because endpoints are inclusive, so a process still runs at its end minute. Sort the events by time, and at equal times process the -1 before the +1 (or just sort tuples, since -1 < +1). Walk through, keep a running count, and track the max. That's O(n log n). The common pitfall is treating endTime as exclusive, which undercounts when one interval ends at minute 3 and another starts at minute 3. Example 1 catches this: the answer is 3 only because the endpoints are inclusive. The heap approach also works: sort by start, pop ends less than the current start, push the new end, track the heap size. If the pattern slips away under pressure, StealthCoder can hand you the sweep during the live OA.

The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.