merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

MergeIntervals.kt

/*
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.
* */

fun merge(intervals: Array<IntArray>): Array<IntArray> {
    if (intervals.size <= 1) {
        return intervals
    }

    // first element determines the start of the interval
    // later, it'll be used to determine when to create a new result item
    intervals.sortBy { it.first() } // here's the time complexity, O(nlogn)
    val result: MutableList<IntArray> = mutableListOf() // necessary, minimal space complexity to return result
    var head: IntArray = intervals.first() // the interval we're editing the end value

    for (i in 1 until intervals.size) {

        val curr: IntArray = intervals[i]

        if (head.last() >= curr.first()) { // grow interval
            head[1] = Math.max(head.last(), curr.last())
        } else { // create new interval
            result.add(head)
            head = curr
        }
    }

    result.add(head) // don't forget the greatest one

    return result.toTypedArray()
}