Start by sorting the intervals by their start times, then iterate through them while maintaining a result list. For each interval, if it overlaps or touches the last interval in the result (i.e., its start <= last end), merge them by updating the end to the maximum of both ends; otherwise, append it. This yields O(n log n) time due to sorting and O(n) space for the output.
Pro tip: Clarify the definition of 'touching' (e.g., [1,2] and [2,3] should merge) and confirm whether the input can be empty or contain invalid intervals; handling these edge cases upfront shows attention to detail and prevents bugs.
Ask about edge cases: empty list, single interval, intervals with start > end, and whether touching intervals should merge. Confirm the expected output format.
Sort the intervals by their start times. If starts are equal, sorting by end is optional but can simplify merging.
Initialize an empty result list. For each interval, if the result is empty or the current interval's start is greater than the last merged interval's end, append it; otherwise, merge by updating the last interval's end to the maximum of both ends.
After processing all intervals, return the merged list, which is already sorted due to the initial sort.
State that time complexity is O(n log n) due to sorting, and space complexity is O(n) for the output (or O(1) extra if merging in-place is allowed).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.