|
| 1 | +from typing import List |
| 2 | + |
| 3 | + |
| 4 | +def merge_intervals(intervals: List[List[int]]) -> List[List[int]]: |
| 5 | + """ |
| 6 | + Merge all overlapping intervals. |
| 7 | +
|
| 8 | + Each interval is represented as a list of two integers [start, end]. |
| 9 | + The function merges overlapping intervals and returns a list of |
| 10 | + non-overlapping intervals sorted by start time. |
| 11 | +
|
| 12 | + Parameters: |
| 13 | + intervals (List[List[int]]): A list of intervals. |
| 14 | +
|
| 15 | + Returns: |
| 16 | + List[List[int]]: A list of merged non-overlapping intervals. |
| 17 | +
|
| 18 | + Examples: |
| 19 | + >>> merge_intervals([[1, 3], [2, 6], [8, 10], [15, 18]]) |
| 20 | + [[1, 6], [8, 10], [15, 18]] |
| 21 | + >>> merge_intervals([[1, 4], [4, 5]]) |
| 22 | + [[1, 5]] |
| 23 | + >>> merge_intervals([[6, 8], [1, 3], [2, 4]]) |
| 24 | + [[1, 4], [6, 8]] |
| 25 | + >>> merge_intervals([]) |
| 26 | + [] |
| 27 | + >>> merge_intervals([[1, 4]]) |
| 28 | + [[1, 4]] |
| 29 | +
|
| 30 | + Time Complexity: |
| 31 | + O(n log n), where n is the number of intervals (sorting step). |
| 32 | +
|
| 33 | + Space Complexity: |
| 34 | + O(n), for storing the merged intervals. |
| 35 | + """ |
| 36 | + |
| 37 | + if not intervals: |
| 38 | + return [] |
| 39 | + |
| 40 | + # Sort intervals based on the start time |
| 41 | + intervals.sort(key=lambda interval: interval[0]) |
| 42 | + |
| 43 | + merged: List[List[int]] = [intervals[0]] |
| 44 | + |
| 45 | + for current in intervals[1:]: |
| 46 | + last = merged[-1] |
| 47 | + |
| 48 | + # If current interval overlaps with the last merged interval |
| 49 | + if current[0] <= last[1]: |
| 50 | + last[1] = max(last[1], current[1]) |
| 51 | + else: |
| 52 | + merged.append(current) |
| 53 | + |
| 54 | + return merged |
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