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 <= 105intervals[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
}
}