# 2580. Count Ways to Group Overlapping Ranges

## Description

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

## Solutions

Solution 1: Sorting + Counting + Fast Power

We can first sort the intervals in the range, merge the overlapping intervals, and count the number of non-overlapping intervals, denoted as $cnt$.

Each non-overlapping interval can be chosen to be put in the first group or the second group, so the number of plans is $2^{cnt}$. Note that $2^{cnt}$ may be very large, so we need to take modulo $10^9 + 7$. Here, we can use fast power to solve this problem.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the number of intervals.

Alternatively, we can also avoid using fast power. Once a new non-overlapping interval is found, we multiply the number of plans by 2 and take modulo $10^9 + 7$.

• class Solution {
public int countWays(int[][] ranges) {
Arrays.sort(ranges, (a, b) -> a[0] - b[0]);
int cnt = 0, mx = -1;
for (int[] e : ranges) {
if (e[0] > mx) {
++cnt;
}
mx = Math.max(mx, e[1]);
}
return qpow(2, cnt, (int) 1e9 + 7);
}

private int qpow(long a, int n, int mod) {
long ans = 1;
for (; n > 0; n >>= 1) {
if ((n & 1) == 1) {
ans = ans * a % mod;
}
a = a * a % mod;
}
return (int) ans;
}
}

• class Solution {
public:
int countWays(vector<vector<int>>& ranges) {
sort(ranges.begin(), ranges.end());
int cnt = 0, mx = -1;
for (auto& e : ranges) {
cnt += e[0] > mx;
mx = max(mx, e[1]);
}
using ll = long long;
auto qpow = [&](ll a, int n, int mod) {
ll ans = 1;
for (; n; n >>= 1) {
if (n & 1) {
ans = ans * a % mod;
}
a = a * a % mod;
}
return ans;
};
return qpow(2, cnt, 1e9 + 7);
}
};

• class Solution:
def countWays(self, ranges: List[List[int]]) -> int:
ranges.sort()
cnt, mx = 0, -1
for start, end in ranges:
if start > mx:
cnt += 1
mx = max(mx, end)
mod = 10**9 + 7
return pow(2, cnt, mod)


• func countWays(ranges [][]int) int {
sort.Slice(ranges, func(i, j int) bool { return ranges[i][0] < ranges[j][0] })
cnt, mx := 0, -1
for _, e := range ranges {
if e[0] > mx {
cnt++
}
if mx < e[1] {
mx = e[1]
}
}
qpow := func(a, n, mod int) int {
ans := 1
for ; n > 0; n >>= 1 {
if n&1 == 1 {
ans = ans * a % mod
}
a = a * a % mod
}
return ans
}
return qpow(2, cnt, 1e9+7)
}

• function countWays(ranges: number[][]): number {
ranges.sort((a, b) => a[0] - b[0]);
let mx = -1;
let ans = 1;
const mod = 10 ** 9 + 7;
for (const [start, end] of ranges) {
if (start > mx) {
ans = (ans * 2) % mod;
}
mx = Math.max(mx, end);
}
return ans;
}