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2580. Count Ways to Group Overlapping Ranges
Description
You are given a 2D integer array ranges
where ranges[i] = [start_{i}, end_{i}]
denotes that all integers between start_{i}
and end_{i}
(both inclusive) are contained in the i^{th}
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 because2
and3
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 10^{9} + 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 <= 10^{5}
ranges[i].length == 2
0 <= start_{i} <= end_{i} <= 10^{9}
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 nonoverlapping intervals, denoted as $cnt$.
Each nonoverlapping 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 nonoverlapping 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; }