1. Insert Interval

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

Approach

• We loop through each interval we check for out of ranges first
• First condition is new interval starting ahead of the interval’s end so we add interval since it is behind
• Second conditions is new interval is behind the interval so we add both the new and interval and assign new interval null as this is added now
• Last condition is when we have some overlapping we simply check min and max for start and end respectively
• After the loop it is possible that our new interval is beyond our given array so we check if new interval is added to the solution or not
• Time: O(n) Space: O(1) space for solution is n but cannot be used for measuring as that is the output

class Solution {
    public int[][] insert(int[][] intervals, int[] newInterval) {
        List<int[]> a = new ArrayList<>();
        for (int[] interval: intervals) {
            if (newInterval == null || interval[1] < newInterval[0])
                a.add(interval);
            else if (newInterval[1] < interval[0]) {
                a.add(newInterval);
                a.add(interval);
                newInterval = null;
            } else {
                newInterval[0] = Math.min(newInterval[0],interval[0]);
                newInterval[1] = Math.max(newInterval[1],interval[1]);
            }
        }
        if (newInterval != null)
            a.add(newInterval);
 
        return a.toArray(new int[a.size()][]);    
    }
}