# LeetCode 57. Insert Interval

## Description

https://leetcode.com/problems/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:

```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]`.```

Example 3:

```Input: intervals = [], newInterval = [5,7]
Output: [[5,7]]
```

Example 4:

```Input: intervals = [[1,5]], newInterval = [2,3]
Output: [[1,5]]
```

Example 5:

```Input: intervals = [[1,5]], newInterval = [2,7]
Output: [[1,7]]
```

Constraints:

• `0 <= intervals.length <= 104`
• `intervals[i].length == 2`
• `0 <= intervals[i][0] <= intervals[i][1] <= 105`
• `intervals` is sorted by `intervals[i][0]` in ascending order.
• `newInterval.length == 2`
• `0 <= newInterval[0] <= newInterval[1] <= 105`

## Explanation

We can insert the new interval first then merge all intervals.

## Python Solution

``````class Solution:
def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:

intervals.append(newInterval)

intervals = sorted(intervals, key=lambda interval:interval[0])

results = []

prev = intervals[0]

for interval in intervals[1:]:
if interval[0] <= prev[1]:
prev[1] = max(interval[1], prev[1])
else:
results.append(prev)
prev = interval

results.append(prev)

return results``````
• Time Complexity: O(Nlog(N)).
• Space Complexity: O(N).