2580. Count Ways to Group Overlapping Ranges

Medium

216

14

You are given a 2D integer array `ranges`

where `ranges[i] = [start`

denotes that all integers between _{i}, end_{i}]`start`

and _{i}`end`

(both _{i}**inclusive**) are contained in the `i`

range.^{th}

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`

`10`^{9} + 7

.

**Example 1:**

Input:ranges = [[6,10],[5,15]]Output:2Explanation: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:4Explanation: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 <= 10`

^{5}`ranges[i].length == 2`

`0 <= start`

_{i}<= end_{i}<= 10^{9}

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