text: # 435. Non-overlapping Intervals
  # https://leetcode.com/problems/non-overlapping-intervals
  
  def eraseOverlapIntervals(intervals):
      if not intervals:
          return 0
      intervals.sort()
      count = 0
  
      new_interval = intervals[0]
      for i in range(1, len(intervals)):
          if intervals[i][0] < new_interval[1]:
              count += 1
              new_interval = [min(new_interval[0], intervals[i][0]),
                              min(new_interval[1], intervals[i][1])]
          else:
              new_interval = intervals[i]
      return count
  
  
  
  
  list = [[1,2],[2,3],[3,4],[1,3]]
  ans = eraseOverlapIntervals(list)
  print(ans)
  
  
  
  # Example 1:
  #
  # Input: intervals = [[1,2],[2,3],[3,4],[1,3]]
  # Output: 1
  # Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping.
  # Example 2:
  #
  # Input: intervals = [[1,2],[1,2],[1,2]]
  # Output: 2
  # Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.
  # Example 3:
  #
  # Input: intervals = [[1,2],[2,3]]
  # Output: 0
  # Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
  