0361MediumArray

Slot a New Rehearsal Block

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A rehearsal room stores its booked blocks in intervals. Each block is a pair [from, to] giving the minute the block opens and the minute it closes. The blocks are listed from earliest opening to latest, and no two of them share even a single minute, not even an endpoint.

A fresh request arrives as newInterval, another [from, to] pair. Put it on the schedule. Every booked block the request runs into fuses with it into one longer block, and blocks that merely meet the request at an endpoint count as running into it.

Return the schedule after the request has been placed: a list of blocks ordered by opening minute, with no two blocks overlapping or meeting.

Examples

Example 1

Input: intervals = [[1,3],[6,9],[12,14]], newInterval = [2,7]
Output: [[1, 9], [12, 14]]

The request covers minutes 2 through 7, so it runs into [1,3] and into [6,9]; those two and the request fuse into [1,9]. The block [12,14] is left alone.

Example 2

Input: intervals = [[5,8]], newInterval = [2,5]
Output: [[2, 8]]

The request closes on the same minute the booked block opens, and meeting at an endpoint counts, so the two fuse into [2,8].

Example 3

Input: intervals = [], newInterval = [4,9]
Output: [[4, 9]]

Nothing is booked, so the schedule is the request on its own.

Example 4

Input: intervals = [[1,3],[5,6],[8,10],[14,16]], newInterval = [4,9]
Output: [[1, 3], [4, 10], [14, 16]]

The request runs into [5,6] and [8,10], and the three fuse into [4,10]. It opens at minute 4 while [1,3] closes at minute 3, so those two neither overlap nor meet, and [14,16] is untouched.

Constraints

0 <= intervals.length <= 10^4
intervals[i].length == 2
0 <= intervals[i][0] <= intervals[i][1] <= 10^5
intervals is ordered by opening minute, and no two blocks overlap or meet.
newInterval.length == 2
0 <= newInterval[0] <= newInterval[1] <= 10^5

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def insert(intervals: list[list[int]], newInterval: list[int]) -> list[list[int]]:

Java

public int[][] insert(int[][] intervals, int[] newInterval)

The intended approach

Time O(n)Space O(n) for the returned schedule, O(1) beyond it

Because the schedule is already ordered and separated, one pass in three stages is enough. Copy across every block that closes before the request opens, then absorb every following block that opens no later than the running close minute, widening the request as you go, then emit the widened block and copy the rest across unchanged.

Pythonentry point: insert

def insert(intervals, newInterval):
    # One left-to-right pass. O(n) time, O(n) for the schedule that is returned.
    out = []
    start, end = newInterval
    i, n = 0, len(intervals)
    while i < n and intervals[i][1] < start:
        out.append(intervals[i])
        i += 1
    while i < n and intervals[i][0] <= end:
        if intervals[i][0] < start:
            start = intervals[i][0]
        if intervals[i][1] > end:
            end = intervals[i][1]
        i += 1
    out.append([start, end])
    while i < n:
        out.append(intervals[i])
        i += 1
    return out

The obvious version

Throw the request in with the booked blocks, sort the whole lot by opening minute, then sweep once fusing each block into the previous one whenever they touch or overlap.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_insert(intervals, newInterval):
    blocks = sorted(list(intervals) + [list(newInterval)])
    merged = []
    for block in blocks:
        if merged and block[0] <= merged[-1][1]:
            if block[1] > merged[-1][1]:
                merged[-1][1] = block[1]
        else:
            merged.append([block[0], block[1]])
    return merged

How this problem was checked

Reference solution executed against 11 test cases — 11 passed
4 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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