57. Insert Interval
Tags: java, algorithms, leetcode, 100daysofcode
Published: 2023-01-16 04:27:52

**Problem Statement:**

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.

 

**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

**Solution:**
**Algorithm:**
1. Create a new array of size intervals.length + 1. This will be the final result.
2. Iterate through the intervals array. If the end of the current interval is less than the start of the new interval, then add the current interval to the result array and increment the index of the result array.
3. If the start of the current interval is less than or equal to the end of the new interval, then merge the current interval with the new interval. The start of the new interval is the minimum of the start of the current interval and the start of the new interval. The end of the new interval is the maximum of the end of the current interval and the end of the new interval. Increment the index of the intervals array.
4. Add the new interval to the result array and increment the index of the result array.
5. Add the remaining intervals to the result array.
6. Create a new array of size j. This will be the final result.
7. Copy the elements from the result array to the final result array.
8. Return the final result array.

**Code:**
```
public class Solution {
    public int[][] insert(int[][] intervals, int[] newInterval) {
        int[][] result = new int[intervals.length + 1][2];
        int i = 0;
        int j = 0;
        while(i < intervals.length && intervals[i][1] < newInterval[0]){
            result[j] = intervals[i];
            i++;
            j++;
        }
        while(i < intervals.length && intervals[i][0] <= newInterval[1]){
            newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
            newInterval[1] = Math.max(newInterval[1], intervals[i][1]);
            i++;
        }
        result[j] = newInterval;
        j++;
        while(i < intervals.length){
            result[j] = intervals[i];
            i++;
            j++;
        }
        int[][] finalResult = new int[j][2];
        for(int k = 0; k < j; k++){
            finalResult[k] = result[k];
        }
        return finalResult;

    }
    
}
```

**Time Complexity:**
O(N), where N is the number of intervals.

**Space Complexity:**
O(N), where N is the number of intervals.