Question

Problem description is follows:

There are n events for particular day d having start time and duration. Example:

e1 10:15:06 11ms (ms = milli seconds)
e2 10:16:07 12ms
......

I need to find out the time x and n. Where x is the time when maximum events were getting executed.

Solution I am thinking is:
Scanning all ms in day d. But that request total 86400000*n calculation. Example

Check at 00::00::00::001 How many events are running
Check at 00::00::00::002 How many events are running
Take max of Range(00::00::00::01,00::00::00::00)

Second solution I am thinking is:

For eventi in all events
   Set running_event=1
   eventj in all events Where eventj!=eventi
        if eventj.start_time in Range (eventi.start_time,eventi.execution_time)
           running_event++

And then take max of running_event

Is there any better solution for this?

Answer 1

This can be solved in O(n log n) time:

• Make an array of all events. This array is already partially sorted: O(n)
• Sort the array: O(n log n); your library should be able to make use of the partial sortedness (timSort does that very well); look into distribution-based sorting algorithms for better expected running time.
  • Sort event boundaries ascending w.r.t. the boundary time
  • Sort event ends before sort starts if touching intervals are considered non-overlapping
    (Sort event ends after sort starts if touching intervals are considered overlapping)
• Initialise running = 0, running_best = 0, best_at = 0
• For each event boundary:
  • If it’s a start of an event, increment running
  • If running > running_best, set best_at = current event time
  • If it’s an end of an event, decrement running
• output best_at