[LC] 56. Merge Intervals
TechGiven 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.