Package g2501_2600.s2580_count_ways_to_group_overlapping_ranges

Class Solution

java.lang.Object
g2501_2600.s2580_count_ways_to_group_overlapping_ranges.Solution
public class Solution extends Object
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

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • countWays

      public int countWays(int[][] ranges)