How do you solve Merge Intervals on LeetCode?

Merge Intervals (LeetCode #56) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting helps in efficiently merging overlapping intervals.

What is the brute force solution for Merge Intervals?

The brute force approach for Merge Intervals is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every pair of intervals to see if they overlap and merging them if they do. This method is straightforward but inefficient for larger datasets. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Merge Intervals?

Merge Intervals uses the following concepts: Array, Sorting. The recommended patterns to study are: Sorting, Greedy Algorithms.

What are the interview tips for Merge Intervals?

Always clarify the problem statement and constraints before jumping to code. Think about edge cases and how your solution handles them. Explain your thought process clearly as you code, as this shows your understanding.

What are common mistakes when solving Merge Intervals?

Not considering edge cases like empty input or single interval. Failing to update the last merged interval correctly.

#56

Merge Intervals

Medium

ArraySortingSortingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution sorts the intervals first and then merges them in a single pass. This is efficient because once sorted, overlapping intervals will be adjacent, allowing us to merge them easily.

⚙️

Algorithm

3 steps

1Step 1: Sort the intervals by their starting times.
2Step 2: Initialize an empty list for merged intervals and add the first interval.
3Step 3: Iterate through the sorted intervals, merging them if they overlap with the last added interval.

solution.py11 lines

1def merge(intervals):
2    if not intervals:
3        return []
4    intervals.sort(key=lambda x: x[0])
5    merged = [intervals[0]]
6    for i in range(1, len(intervals)):
7        if merged[-1][1] >= intervals[i][0]:
8            merged[-1][1] = max(merged[-1][1], intervals[i][1])
9        else:
10            merged.append(intervals[i])
11    return merged

ℹ

Complexity note: The sorting step takes O(n log n) time, and the merging step takes O(n) time, leading to an overall time complexity of O(n log n). Space complexity is O(n) due to the storage of merged intervals.

1Sorting helps in efficiently merging overlapping intervals.
2Overlapping intervals will always be adjacent after sorting.

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