How do you solve Insert Interval on LeetCode?

Insert Interval (LeetCode #57) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The intervals are sorted, which allows for efficient merging.

What is the time complexity of Insert Interval?

The optimal solution for Insert Interval runs in O(n) time and uses O(n) space. The complexity is O(n) because we only make a single pass through the intervals, and O(n) space is used for the result list.

What is the brute force solution for Insert Interval?

The brute force approach for Insert Interval is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves simply adding the new interval to the list and then merging any overlapping intervals. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Insert Interval?

Insert Interval uses the following concepts: Array. The recommended patterns to study are: Two Pointers, Array.

What are the interview tips for Insert Interval?

Always check if the input list is empty before processing. Think about edge cases, like inserting at the beginning or end. Practice merging intervals separately to build confidence.

What are common mistakes when solving Insert Interval?

Not considering the case where the new interval is completely before or after all existing intervals. Failing to merge overlapping intervals correctly.

#57

Insert Interval

Medium

ArrayTwo PointersArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach leverages the sorted nature of the intervals to efficiently insert and merge the new interval in a single pass, resulting in better performance.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty list for the result.
2Step 2: Iterate through the existing intervals and add them to the result until you find an interval that overlaps with the new interval.
3Step 3: Merge the overlapping intervals with the new interval.
4Step 4: Add any remaining intervals to the result.

solution.py13 lines

1def insert(intervals, newInterval):
2    merged = []
3    i = 0
4    while i < len(intervals) and intervals[i][1] < newInterval[0]:
5        merged.append(intervals[i])
6        i += 1
7    while i < len(intervals) and intervals[i][0] <= newInterval[1]:
8        newInterval[0] = min(newInterval[0], intervals[i][0])
9        newInterval[1] = max(newInterval[1], intervals[i][1])
10        i += 1
11    merged.append(newInterval)
12    merged.extend(intervals[i:])
13    return merged

ℹ

Complexity note: The complexity is O(n) because we only make a single pass through the intervals, and O(n) space is used for the result list.

1The intervals are sorted, which allows for efficient merging.
2Merging can be done in a single pass if we handle overlaps correctly.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.