56_Merge Intervals

[Medium][Array, Sort]

Given a collection of intervals, merge all overlapping intervals.

Example 1:

Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:

Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

Solution:

Idea:

• Sort the intervals by the starting number.

• Compare if next starting number is less than or equal to previous ending number.

  • If previous ending >= next starting, merge together

  • If previous ending < next starting, keep splitted.

Time Complexity: $O(n\log{n})$

Space Complexity: $O(1)$

# Definition for an interval.
# class Interval(object):
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

def merge(intervals):
    """
    :type intervals: List[Interval]
    :rtype: List[Interval]
    """
    if len(intervals) == 0:
        return []
    intervals.sort(key = lambda x: x.start)
    output = [intervals[0]]
    for n in intervals[1:]:
        if output[-1].end >= n.start:
            output[-1].end = max(n.end, output[-1].end)
        else:
            output.append(n)
    return output