LeetCode in Kotlin

435. Non-overlapping Intervals

Medium

Given an array of intervals intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Example 1:

Input: intervals = [[1,2],[2,3],[3,4],[1,3]]

Output: 1

Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping.

Example 2:

Input: intervals = [[1,2],[1,2],[1,2]]

Output: 2

Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.

Example 3:

Input: intervals = [[1,2],[2,3]]

Output: 0

Explanation: You don’t need to remove any of the intervals since they’re already non-overlapping.

Constraints:

• 1 <= intervals.length <= 105
• intervals[i].length == 2
• -5 * 104 <= starti < endi <= 5 * 104

Solution

class Solution {
    fun eraseOverlapIntervals(intervals: Array<IntArray>): Int {
        intervals.sortWith { a: IntArray, b: IntArray ->
            if (a[0] != b[0]
            ) {
                a[0] - b[0]
            } else {
                a[1] - b[1]
            }
        }
        var erasures = 0
        var end = intervals[0][1]
        for (i in 1 until intervals.size) {
            end = if (intervals[i][0] < end) {
                erasures++
                Math.min(end, intervals[i][1])
            } else {
                intervals[i][1]
            }
        }
        return erasures
    }
}

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