old_contents: # Time:  O(n)
  # Space: O(1)
  #
  # Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
  # 
  # You may assume that the intervals were initially sorted according to their start times.
  # 
  # Example 1:
  # Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
  # 
  # Example 2:
  # Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].
  # 
  # This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].
  #
  
  # Definition for an interval.
  class Interval:
      def __init__(self, s=0, e=0):
          self.start = s
          self.end = e
          
      def __repr__(self):
          return "[{}, {}]".format(self.start, self.end)
  
  class Solution:
      # @param intervals, a list of Intervals
      # @param newInterval, a Interval
      # @return a list of Interval
      def insert(self, intervals, newInterval):
          return self.merge(intervals + [newInterval])
          
      def merge(self, intervals):
          if not intervals:
              return intervals
          intervals.sort(key = lambda x: x.start)
          result = [intervals[0]]
          for i in xrange(1, len(intervals)):
              prev, current = result[-1], intervals[i]
              if current.start <= prev.end: 
                  prev.end = max(prev.end, current.end)
              else:
                  result.append(current)
          return result
  
  if __name__ == "__main__":
      print Solution().insert([Interval(1, 2), Interval(3, 5), Interval(6, 7), Interval(8, 10), Interval(12, 16)], Interval(4, 9))
  
new_contents: # Time:  O(n)
  # Space: O(1)
  
  # Given a set of non-overlapping intervals, insert a new interval into the
  # intervals (merge if necessary).
  # You may assume that the intervals were initially sorted according to their start times.
  # 
  # Example 1:
  # Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
  # 
  # Example 2:
  # Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].
  # 
  # This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].
  
  # Definition for an interval.
  class Interval:
      def __init__(self, s=0, e=0):
          self.start = s
          self.end = e
          
      def __repr__(self):
          return "[{}, {}]".format(self.start, self.end)
  
  
  class Solution(object):
      def insert(self, intervals, newInterval):
          """
          :type intervals: List[Interval]
          :type newInterval: Interval
          :rtype: List[Interval]
          """
          result = []
          i = 0
          while i < len(intervals) and newInterval.start > intervals[i].end:
              result += intervals[i],
              i += 1
          while i < len(intervals) and newInterval.end >= intervals[i].start:
              newInterval = Interval(min(newInterval.start, intervals[i].start), \
                                     max(newInterval.end, intervals[i].end))
              i += 1
          result += newInterval,
          result += intervals[i:]
          return result
  
  
  if __name__ == "__main__":
      print Solution().insert([Interval(1, 2), Interval(3, 5), Interval(6, 7), Interval(8, 10), Interval(12, 16)], Interval(4, 9))
  
subject: Update insert-interval.py