[LC] 56. Merge Intervals

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] overlaps, 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.

Constraints:

• 1 <= intervals.length <= 104
• intervals[i].length == 2
• 0 <= starti <= endi <= 104

Approach

Sort by 1st element, then check if the next interval is overlapped. If the checking interval’s 1st element is more than the current end time, then change the checking interval to the current interval.

class Solution:
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
        rslt = []
        if len(intervals) == 0:
            return rslt
        
        intervals.sort()
        start_x, start_y = intervals[0]
        
        for i in range(1, len(intervals)):
            if start_y < intervals[i][0]:
                rslt.append([start_x, start_y])
                start_x, start_y = intervals[i]
            elif start_y < intervals[i][1]:
                start_y = intervals[i][1]
                
        rslt.append([start_x, start_y])
        return rslt
                

Time Complexity: O(NlogN)
Space Complexity: O(1)

Optimization: maybe we can use the heap?

Submission

Runtime: 92 ms, faster than 35.73% of Python3 online submissions for Merge Intervals.
Memory Usage: 16.3 MB, less than 12.13% of Python3 online submissions for Merge Intervals.