57.Insert_Interval_(Medium).md

57. Insert Interval (Medium)

Update: Jun 6, 2024, 6:17 PM (EST)

Link: https://leetcode.com/problems/insert-interval/

Question:

You are given an array of non-overlapping intervals intervals where intervals[i] = [start_i</sub>, end_i] represent the start and the end of the ith interval and intervals is sorted in ascending order by start_i. 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 start_i 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].

Confusion:

While we are comparing intervals, it is one interval from intervals to compare with newInterval, e.g. [1, 2] vs [3,5], not a list of intervals to compare with newInterval: [[1, 2], [4, 5]] vs [3, 6]. So we should only consider the cases of one vs one, ie. the cases when lst[1] < newInterval[0] or newInterval[1] < lst[0].

My Solution:

1. interval: [4, 8] < newInterval: [9, 11], interval[1] < newInterval[0], the interval is before newInterval, so we add interval to our created list.

2. interval: [5, 6] > newInterval: [3, 4], inverval[0] > newInterval[1], the newInterval is before interval, so we add the newInterval to the list and append all the rest intervals intervals[i:] to the created list.

3. interval: [1, 4] < newInterval: [3, 5], this is the case when overlap happens (newInterval[0] < interval[1]). To handle the overlap, we iteratively update the newInterval by updating min between newInterval[0] and interval[0], max between newInterval[1] and interval[1]. When the newInterval has no more overlap, it has to satisfy the first two base cases and return. If not, this is the last case when we only need to add the newInterval to the end of the created list: intervals = [], newInterval = [5, 7].

So we just append newInterval to the created list after the for loop ends.

class Solution:
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        newLst = []
        for i in range(len(intervals)):
            # interval = [1, 2], newInterval = [3, 5], 2 < 3
            if intervals[i][1] < newInterval[0]:
                newLst.append(intervals[i])
            # interval = [6, 8], newInterval = [3, 5], 5 < 6
            elif newInterval[1] < intervals[i][0]:
                newLst.append(newInterval)
                return newLst + intervals[i:]
            else:   # Else case when overlapping happen
                newInterval = [min(intervals[i][0], newInterval[0]), max(intervals[i][1], newInterval[1])]
        # When newInterval should be added to the end of newLst or intervals = []
        newLst.append(newInterval)
        return newLst