2580

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

Companies: Oracle

Related Topics:
Array, Sorting

Similar Questions:

• Merge Intervals (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/count-ways-to-group-overlapping-ranges
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int countWays(vector<vector<int>>& A) {
        sort(begin(A), end(A));
        long end = INT_MIN, mod = 1e9 + 7, cnt = 0;
        for (auto &r : A) {
            cnt += r[0] > end;
            end = max(end, (long)r[1]);
        }
        auto pow = [&](long base, long e) {
            long ans = 1;
            while (e) {
                if (e & 1) ans = ans * base % mod;
                base = base * base % mod;
                e >>= 1;
            }
            return ans;
        };
        return pow(2, cnt);
    }
};