0057-insert-interval

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Description

You are given an array of non-overlapping intervals intervals where intervals[i] = [starti, endi] represent the start and the end of the ith interval and intervals is sorted in ascending order by starti. You are also given an interval newInterval = [start, end] that represents the start and end of another interval.

Insert newInterval into intervals such that intervals is still sorted in ascending order by starti and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary).

Return intervals after the insertion.

Note that you don't need to modify intervals in-place. You can make a new array and return it.

 

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].

 

Constraints:

  • 0 <= intervals.length <= 104
  • intervals[i].length == 2
  • 0 <= starti <= endi <= 105
  • intervals is sorted by starti in ascending order.
  • newInterval.length == 2
  • 0 <= start <= end <= 105

Solution(Python)

class Solution:
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        # Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
        # Output: [[1,5],[6,9]]
        # Example 2:
        # [[1,3],[6,9]]
        #   2 < 3   add to 1,3
        #   insert_point = binary search 
        #   merge if overlap
        #   insert 
        # case 1  NO overlap before merging 
        #   before overlap    new interval start >  interval i end
        #   # 2 overlap and mergubf
        #        during iverlap    newinterval end >= intervak start
        #  # 3 no pverlap after merging
        # n = len(intervals)
        # i = 0
        # res = []
        # while i<n and intervals[i][1] <   newInterval[0]:
        #     res.append(intervals[i])
        #     i += 1
        # while i<n and intervals[i][0] <=   newInterval[1]:
        #     newInterval[0] = min(newInterval[0], intervals[i][0])
        #     newInterval[1] = max(newInterval[1], intervals[i][1])
        #     i += 1
        # res.append(newInterval)
        # while i<n:
        #     res.append(intervals[i])
        #     i += 1
        # return res
        if not intervals:
            return [newInterval]
        
        n = len(intervals)

        target = newInterval[0]
        left, right = 0, n - 1


        while left <= right:
            mid = (left + right) // 2
            if intervals[mid][0] < target:
                left = mid + 1
            else:
                right = mid - 1
        intervals.insert(left, newInterval)
        res = []

        for interval in intervals:
            # If res is empty or there is no overlap, add the interval to the result
            if not res or res[-1][1] < interval[0]:
                res.append(interval)
            # If there is an overlap, merge the intervals by updating the end of the last interval in res
            else:
                res[-1][1] = max(res[-1][1], interval[1])
        return res

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