#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/greedy-algorithms/interval-problems/non-overlapping-implementation

░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░███████████████████████████████████████████████████████████████████████████████████████████████

#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/greedy-algorithms/interval-problems/non-overlapping-implementation

Repovive

Non-overlapping - Implementation - Interval Problems | Greedy Algorithms | Repovive

Repovive.

Greedy Algorithms8 sections · 317 units14h total

Open in Course

Non-overlapping - Implementation

The code

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.