LeetCode-in-All

56. Merge Intervals

Medium

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] overlap, 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

Solution

-spec merge(Intervals :: [[integer()]]) -> [[integer()]].
merge(Intervals) ->
    SortedIntervals = lists:sort(fun([Start1, _], [Start2, _]) -> Start1 =< Start2 end, Intervals),
    lists:reverse(
        lists:foldl(fun([Start, Finish], Acc) ->
            case Acc of
                [] -> [[Start, Finish]];
                [[PrevStart, PrevFinish] | Rest] ->
                    if PrevFinish >= Start ->
                        %% Merge overlapping intervals
                        [[PrevStart, erlang:max(PrevFinish, Finish)] | Rest];
                       true ->
                        %% No overlap
                        [[Start, Finish] | Acc]
                    end
            end
        end, [], SortedIntervals)
    ).

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