---
title: '56. Merge Intervals'
description: Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input
icon: dot
topics:
  - { name: "Array", slug: "array" }
  - { name: "Sorting", slug: "sorting" }
  - { name: "Quicksort", slug: "quicksort" }
issue: "https://github.com/prdlk/leetcode/issues/182"
sidebar:
  label: 'Merge Intervals'
  badge: 'Medium'
---

### Example 1:
- Input: `intervals = [[1,3],[2,6],[8,10],[15,18]]`
- Output: `[[1,6],[8,10],[15,18]]`
- Explanation: Since `intervals [1,3]` and `[2,6]` overlap, merge them into `[1,6]`.

### Example 2:
- Input: `intervals = [[1,4],[4,5]]`
- Output: `[[1,5]]`
- Explanation: Intervals `[1,4]` and `[4,5]` are considered overlapping.

### Example 3:
- Input: `intervals = [[4,7],[1,4]]`
- Output: `[[1,7]]`
- Explanation: Intervals `[1,4]` and `[4,7]` are considered overlapping.

### Constraints:

- `1 <= intervals.length <= 10^4`
- `intervals[i].length == 2`
- `0 <= starti <= endi <= 10^4`

## Solution

```py
class Solution:
    def merge(self, intervals: list[list[int]]) -> list[list[int]]:
        intervals.sort(key=lambda x: x[0])
        result = []

        for start, end in intervals:
            if result and start <= result[-1][1]:
                result[-1][1] = max(result[-1][1], end)
            else:
                result.append([start, end])

        return result
```