Non-overlapping - Implementation

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

Time: $O(n \log n)$. Space: $O(1)$ extra.