LeetCode in Kotlin

2580. Count Ways to Group Overlapping Ranges

Medium

You are given a 2D integer array ranges where ranges[i] = [starti, endi] denotes that all integers between starti and endi (both inclusive) are contained in the ith range.

You are to split ranges into two (possibly empty) groups such that:

• Each range belongs to exactly one group.
• Any two overlapping ranges must belong to the same group.

Two ranges are said to be overlapping if there exists at least one integer that is present in both ranges.

• For example, [1, 3] and [2, 5] are overlapping because 2 and 3 occur in both ranges.

Return the total number of ways to split ranges into two groups. Since the answer may be very large, return it modulo 109 + 7.

Example 1:

Input: ranges = [[6,10],[5,15]]

Output: 2

Explanation:

The two ranges are overlapping, so they must be in the same group.

Thus, there are two possible ways:

• Put both the ranges together in group 1.

• Put both the ranges together in group 2.

Example 2:

Input: ranges = [[1,3],[10,20],[2,5],[4,8]]

Output: 4

Explanation:

Ranges [1,3], and [2,5] are overlapping. So, they must be in the same group.

Again, ranges [2,5] and [4,8] are also overlapping. So, they must also be in the same group.

Thus, there are four possible ways to group them:

• All the ranges in group 1.

• All the ranges in group 2.

• Ranges [1,3], [2,5], and [4,8] in group 1 and [10,20] in group 2.

• Ranges [1,3], [2,5], and [4,8] in group 2 and [10,20] in group 1.

Constraints:

• 1 <= ranges.length <= 105
• ranges[i].length == 2
• 0 <= starti <= endi <= 109

Solution

import java.util.Arrays

@Suppress("NAME_SHADOWING")
class Solution {
    fun countWays(ranges: Array<IntArray>): Int {
        var cnt = 1
        Arrays.sort(ranges) { a, b -> if (a[0] != b[0]) a[0] - b[0] else a[1] - b[1] }
        var curr = ranges[0]
        for (i in 1 until ranges.size) {
            if (ranges[i][1] < curr[0] || ranges[i][0] > curr[1]) {
                ++cnt
                curr = ranges[i]
            } else {
                curr[1] = Math.max(curr[1], ranges[i][1])
            }
        }
        return powMod(2, cnt.toLong()).toInt()
    }

    private fun powMod(b: Long, e: Long): Long {
        var b = b
        var e = e
        var ans: Long = 1
        while (e != 0L) {
            if (e and 1L == 1L) {
                ans *= b
                ans %= MOD.toLong()
            }
            b *= b
            b %= MOD.toLong()
            e = e shr 1
        }
        return ans
    }

    companion object {
        var MOD = 1e9.toInt() + 7
    }
}

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