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 because2and3occur 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 <= 105ranges[i].length == 20 <= starti <= endi <= 109
Solution
Python3
class Solution:
def countWays(self, ranges: List[List[int]]) -> int:
N = len(ranges)
ranges.sort()
M = 10 ** 9 + 7
count = 1
right = ranges[0][1]
for i in range(1, N):
if ranges[i][0] > right:
right = ranges[i][1]
count += 1
else:
right = max(right, ranges[i][1])
return pow(2, count, M)