## Problem Statement

Given a list of non-overlapping intervals sorted by their start time, **insert a given interval at the correct position** and merge all necessary intervals to produce a list that has only mutually exclusive intervals.

**Example 1**:

```
Input: Intervals=[[1,3], [5,7], [8,12]], New Interval=[4,6]
Output: [[1,3], [4,7], [8,12]]
Explanation: After insertion, since [4,6] overlaps with [5,7], we merged them into one [4,7].
```

**Example 2**:

```
Input: Intervals=[[1,3], [5,7], [8,12]], New Interval=[4,10]
Output: [[1,3], [4,12]]
Explanation: After insertion, since [4,10] overlaps with [5,7] & [8,12], we merged them into [4,12].
```

**Example 3**:

```
Input: Intervals=[[2,3],[5,7]], New Interval=[1,4]
Output: [[1,4], [5,7]]
Explanation: After insertion, since [1,4] overlaps with [2,3], we merged them into one [1,4].
```

**Constraints:**

- 1 <= intervals.length <= 10<sup>4<sup>
`intervals[i].length == 2`

- 0 <= start<sub>i</sub> <= end<sub>i</sub> <= 10<sup>5<sup>
`intervals`

is sorted by starti in`ascending`

order.`newInterval.length == 2`

- 0 <= start <= end <= 10<sup>5<sup>

## Solution

If the given list was not sorted, we could have simply appended the new interval to it and used the `merge()`

function from `Merge Intervals`

. But since the given list is sorted, we should try to come up with a solution better than O(N * logN).

When inserting a new interval in a sorted list, we need to first find the correct index where the new interval can be placed. In other words, we need to skip all the intervals which end before the start of the new interval. So we can iterate through the given sorted listed of intervals and skip all the intervals with the following condition:

`intervals[i].end < newInterval.start`

Once we have found the correct place, we can follow an approach similar to `Merge Intervals`

to insert and/or merge the new interval. Let’s call the new interval ‘a’ and the first interval with the above condition ‘b’. There are five possibilities:

The diagram above clearly shows the merging approach. To handle all four merging scenarios, we need to do something like this:

`c.start = min(a.start, b.start) c.end = max(a.end, b.end)`

### Step-by-Step Algorithm

- Create an empty list to store the merged intervals.
- Traverse the given list of intervals:
- Add all intervals that end before the new interval starts to the merged list.
- For overlapping intervals:
- Adjust the start of the new interval to the minimum start time.
- Adjust the end of the new interval to the maximum end time.

- Add the merged new interval to the merged list.
- Add all intervals that start after the new interval ends to the merged list.

- Return the merged list of intervals.

#### Algorithm Walkthrough

Using the example Intervals = [[1, 3], [5, 7], [8, 12]], New Interval = [4, 10]:

- Start with an empty list:
`mergedIntervals = []`

- First interval [1, 3]:
- Ends before [4, 10] starts.
- Add to
`mergedIntervals`

:`[[1, 3]]`

- Second interval [5, 7]:
- Overlaps with [4, 10].
- Adjust new interval:
`[4, 10]`

->`[4, 10]`

(no change)

- Third interval [8, 12]:
- Overlaps with [4, 10].
- Adjust new interval:
`[4, 10]`

->`[4, 12]`

- Add merged new interval [4, 12] to
`mergedIntervals`

:`[[1, 3], [4, 12]]`

- No more intervals left to process.
- Final merged intervals:
`[[1, 3], [4, 12]]`

## Code

Here is what our algorithm will look like:

## Time Complexity

As we are iterating through all the intervals only once, the time complexity of the above algorithm is O(N), where ‘N’ is the total number of intervals.

## Space Complexity

Ignoring the space needed for the result list, the algorithm runs in constant space O(1).

If we include the result list, the space complexity will be O(N) as we need to return a list containing all the merged intervals.