LeetCode-in-Java

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:

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

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:

Constraints:

Solution

import java.util.Arrays;

public class Solution {
    private static final int MOD = (int) 1e9 + 7;

    private long powMod(long e) {
        long ans = 1;
        long b = 2;
        while (e != 0) {
            if ((e & 1) == 1) {
                ans *= b;
                ans %= MOD;
            }
            b *= b;
            b %= MOD;
            e >>= 1;
        }
        return ans;
    }

    public int countWays(int[][] ranges) {
        int cnt = 1;
        Arrays.sort(ranges, (a, b) -> a[0] != b[0] ? a[0] - b[0] : a[1] - b[1]);
        int[] curr = ranges[0];
        for (int i = 1; i < ranges.length; i++) {
            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 (int) powMod(cnt);
    }
}