eraseOverlapIntervals.py

'''
435. Non-overlapping Intervals

Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Example 1:
Input: [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.

Example 2:
Input: [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.

Example 3:
Input: [[1,2],[2,3]]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
'''

def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
  if not intervals:
      return 0
  intervals.sort(key= lambda x: x[1])
  # sort the intervals wrt to their eng points
  count = 0
  n = len(intervals)
  prevEnd = intervals[0][1]
  for i in range(1,n):
      if intervals[i][0] < prevEnd:
      # compare with the previous end point
      # if current start is less than previous end point then, the intervals overlap
          count += 1
      else:
          prevEnd = intervals[i][1]
  return count
  
'''
Time complexity = O(nlogn)
Space complexity  = O(1)
'''