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In this approach, we'll first identify where the new interval should be inserted to maintain order. After insertion, we'll iterate over the array and merge overlapping intervals.
Time Complexity: O(n), where n is the number of intervals.
Space Complexity: O(n) due to the allocation for the result.
1using System;
2using System.Collections.Generic;
3public class Solution {
4 public int[][] Insert(int[][] intervals, int[] newInterval) {
5 List<int[]> result = new List<int[]>();
6 int i = 0;
7 while (i < intervals.Length && intervals[i][1] < newInterval[0])
8 result.Add(intervals[i++]);
9 while (i < intervals.Length && intervals[i][0] <= newInterval[1]) {
10 newInterval[0] = Math.Min(newInterval[0], intervals[i][0]);
11 newInterval[1] = Math.Max(newInterval[1], intervals[i][1]);
12 i++;
13 }
14 result.Add(newInterval);
15 while (i < intervals.Length)
16 result.Add(intervals[i++]);
17 return result.ToArray();
18 }
19}C#'s solution mirrors the logic seen in Python and Java, utilizing a List to manage result construction because Lists allow for dynamic resizing unlike standard arrays.
This approach optimizes finding the insert position using binary search. After inserting the new interval, it merges overlapping intervals. This is slightly more efficient when the intervals list is large.
Time Complexity: O(n), even with binary search (O(log n) for insertion, O(n) for merging).
Space Complexity: O(n), to store the new list of intervals.
1using System.Collections.Generic;
public class Solution {
public int[][] Insert(int[][] intervals, int[] newInterval) {
int insertPosition = BinarySearch(intervals, newInterval[0]);
List<int[]> newIntervals = new List<int[]>(intervals);
newIntervals.Insert(insertPosition, newInterval);
List<int[]> mergedIntervals = new List<int[]> { newIntervals[0] };
for (int i = 1; i < newIntervals.Count; i++) {
var lastInterval = mergedIntervals[mergedIntervals.Count - 1];
if (lastInterval[1] >= newIntervals[i][0])
lastInterval[1] = Math.Max(lastInterval[1], newIntervals[i][1]);
else
mergedIntervals.Add(newIntervals[i]);
}
return mergedIntervals.ToArray();
}
private int BinarySearch(int[][] intervals, int start) {
int low = 0, high = intervals.Length;
while (low < high) {
int mid = (low + high) / 2;
if (intervals[mid][0] < start)
low = mid + 1;
else
high = mid;
}
return low;
}
}The C# approach assimilates the binary search method strategically into list operations, simplifying position determination and allowing the residues of the insertion to be effectively handled afterwards.