Non-overlapping - Implementation

Here is the solution:

function eraseOverlapIntervals(intervals)
 if intervals is empty then
 return 0
 sort intervals by end time
 count := 1
 currentEnd := intervals[0].end
 for i from 1 to length - 1
 if intervals[i].start >= currentEnd then
 count := count + 1
 currentEnd := intervals[i].end
 return length - count

Time: $O(n \log n)$ for sorting. Space: $O(1)$ extra if sorting in place.

The key is recognizing that removing minimum equals keeping maximum. Activity selection gives you the maximum.