Insert Interval

MediumArrayIntervals
Category: Greedy
Companies that ask this question:
AmazonFacebookGoogleMicrosoftLinkedIn

Approach

Insert Interval

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.

Examples

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 <= 10^4
  • intervals[i].length == 2
  • 0 <= starti <= endi <= 10^5
  • intervals is sorted by starti in ascending order.
  • newInterval.length == 2
  • 0 <= start <= end <= 10^5

Approach

Key Insight

Process intervals in three phases:

  1. Before: Add all intervals that end before newInterval starts
  2. Merge: Merge all overlapping intervals with newInterval
  3. After: Add all intervals that start after newInterval ends

Solution: Linear Scan

Single pass through intervals.

Algorithm:

  1. Add all intervals ending before newInterval (no overlap)
  2. Merge all intervals that overlap with newInterval
    • Update newInterval start/end to cover all overlaps
  3. Add merged newInterval
  4. Add all remaining intervals (no overlap)
  5. Return result

Complexity Analysis

  • Time Complexity: O(n) - single pass through intervals
  • Space Complexity: O(n) - result array

Pattern Recognition

This problem demonstrates:

  • Interval insertion pattern
  • Linear scan approach
  • Merge logic without sorting
  • Foundation for calendar, scheduling problems
  • Similar to merge intervals but optimized

Solution

java
1import java.util.*;
2
3class Solution {
4    // Solution 1: Linear Scan (Three Phases)
5    public int[][] insert(int[][] intervals, int[] newInterval) {
6        List<int[]> result = new ArrayList<>();
7        int i = 0;
8        int n = intervals.length;
9
10        // Phase 1: Add all intervals ending before newInterval starts
11        while (i < n && intervals[i][1] < newInterval[0]) {
12            result.add(intervals[i]);
13            i++;
14        }
15
16        // Phase 2: Merge all overlapping intervals
17        while (i < n && intervals[i][0] <= newInterval[1]) {
18            // Merge: extend newInterval to cover current interval
19            newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
20            newInterval[1] = Math.max(newInterval[1], intervals[i][1]);
21            i++;
22        }
23
24        // Add merged interval
25        result.add(newInterval);
26
27        // Phase 3: Add all remaining intervals
28        while (i < n) {
29            result.add(intervals[i]);
30            i++;
31        }
32
33        return result.toArray(new int[result.size()][]);
34    }
35
36    // Solution 2: Cleaner Version
37    public int[][] insertClean(int[][] intervals, int[] newInterval) {
38        List<int[]> result = new ArrayList<>();
39        int start = newInterval[0];
40        int end = newInterval[1];
41
42        for (int i = 0; i < intervals.length; i++) {
43            int currStart = intervals[i][0];
44            int currEnd = intervals[i][1];
45
46            // Current interval ends before new interval starts
47            if (currEnd < start) {
48                result.add(intervals[i]);
49            }
50            // Current interval starts after new interval ends
51            else if (currStart > end) {
52                result.add(new int[]{start, end});
53                // Add all remaining intervals
54                for (int j = i; j < intervals.length; j++) {
55                    result.add(intervals[j]);
56                }
57                return result.toArray(new int[result.size()][]);
58            }
59            // Overlapping: merge
60            else {
61                start = Math.min(start, currStart);
62                end = Math.max(end, currEnd);
63            }
64        }
65
66        // Add merged interval (if not already added)
67        result.add(new int[]{start, end});
68        return result.toArray(new int[result.size()][]);
69    }
70}
71
Loading visualizer...