Question

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:

Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]


Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].


Solution

# Definition for an interval.
# class Interval(object):
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

class Solution(object):
def insert(self, intervals, newInterval):
"""
:type intervals: List[Interval]
:type newInterval: Interval
:rtype: List[Interval]
"""
# Naive cases
if len(intervals) == 0:
return [newInterval]
if newInterval.end < intervals[0].start:
return [newInterval] + intervals
if newInterval.start > intervals[-1].end:
return intervals + [newInterval]

res = list()
idx = 0
# Find intervals before the newInterval, non-overlapping
while idx < len(intervals):
if intervals[idx].end < newInterval.start:
res.append(intervals[idx])
idx += 1
else:
break

# Find the start of the overlap
start = min(intervals[idx].start, newInterval.start)
while idx < len(intervals) - 1:
if intervals[idx + 1].start <= newInterval.end:
idx += 1
else:
break

# When the newInterval fits into two intervals
if newInterval.end < intervals[idx].start:
end = newInterval.end
res.append(Interval(start, end))
res.extend(intervals[idx:])
return res
# When overlaps
else:
end = max(intervals[idx].end, newInterval.end)
res.append(Interval(start, end))
res.extend(intervals[idx+1:])
return res