https://leetcode.com/problems/insert-interval/
Insert Interval
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].
Code
# 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]
"""
if not intervals:
return [newInterval]
strt_idx = self.binary_search(intervals, newInterval.start)
end_idx = self.binary_search(intervals, newInterval.end)
if strt_idx ==0 and newInterval.end < intervals[0].start: # Condition 1
intervals[0:1] = [newInterval, intervals[0]]
elif newInterval.start > intervals[strt_idx].end: # Condition 2
intervals[strt_idx:end_idx+1] = [intervals[strt_idx],
Interval(newInterval.start, max(intervals[end_idx].end, newInterval.end))]
else: # Condition 3
intervals[strt_idx:end_idx+1] = [Interval(min(intervals[strt_idx].start, newInterval.start),
max(intervals[end_idx].end, newInterval.end))]
return intervals
def binary_search(self, intervals, t):
left, right = 0, len(intervals)-1
while left <= right:
mid = (left + right) / 2
if t > intervals[mid].start:
left = mid + 1
elif t < intervals[mid].start:
right = mid - 1
else:
return mid
# return index where t >= intervals[index].start
return left-1 if left - 1 >=0 else 0
Idea
Do two rounds of binary search to determine the rightmost interval in `intervals` whose start $latex \leq$ `newInterval.start`. Also determine the rightmost interval whose start $latex \leq$ `newInterval.end`. Then you have three situations to deal with:
