# Merge Intervals

Published: May 24, 2024

Medium Sorting Array

## Problem Description

Given an array of intervals where intervals[i] = [start_i, end_i], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

Constraints:

• 1 <= intervals.length <= 10**4
• intervals[i].length == 2
• 0 <= start_i <= end_i <= 10**4

https://leetcode.com/problems/merge-intervals/description/

## Examples

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
nput: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

## How to Solve

Sort intervals by start time. Compare one by one. If previous end time is less than current start time, add current. If previous end time is greater than or equal to current start time, update previous end time by the current end time.

## Solution

class MergeIntervals {
public:
vector<vector<int>> merge(vector<vector<int>>& intervals) {
sort(intervals.begin(), intervals.end());
vector<vector<int>> result{intervals[0]};
for (int i = 1; i < intervals.size(); ++i) {
if (result.back()[1] >= intervals[i][0]) {
result.back()[1] = max(result.back()[1], intervals[i][1]);
} else {
result.push_back(intervals[i]);
}
}
return result;
}
};

class MergeIntervals:
def merge(self, intervals: 'List[List[int]]') -> 'List[List[int]]':
if len(intervals) < 2: return intervals
intervals.sort(key=lambda x: x[0])
result = [intervals[0]]
for value in intervals[1:]:
if result[-1][1] < value[0]:
result.append(value)
elif result[-1][1] < value[1]:
result[-1][1] = value[1]
return result

## Complexities

• Time: O(n * log(n))
• Space: O(n)