# 56. Merge Intervals

Updated: 2023-03-04

LeetCode problem

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.

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.


Approach 1:

class Solution:
def merge(self, intervals: List[List[int]]) -> List[List[int]]:
intervals.sort()
res = [intervals[0]]
for ir in range(1, len(intervals)):
if intervals[ir][0] >= res[-1][0] and intervals[ir][0] <= res[-1][1]: # [1,3],[2,6]
res[-1][0] = min(intervals[ir][0], res[-1][0])
res[-1][1] = max(intervals[ir][1], res[-1][1])
elif res[-1][0] >= intervals[ir][0] and res[-1][0] <= intervals[ir][1]: # [1,3],[0,4]
res[-1][0] = min(intervals[ir][0], res[-1][0])
res[-1][1] = max(intervals[ir][1], res[-1][1])
else:
res.append(intervals[ir])
return res


Approach 2:

class Solution:
def merge(self, intervals: List[List[int]]) -> List[List[int]]:
ans = []
for interval in sorted(intervals):
if not ans or ans[-1][1] < interval[0]:
ans.append(interval)
else:
ans[-1][1] = max(ans[-1][1], interval[1])
return ans