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:
Two ranges are said to be overlapping if there exists at least one integer that is present in both ranges.
[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
import java.util.Arrays;
public class Solution {
private static final int MOD = (int) 1e9 + 7;
private long powMod(long e) {
long ans = 1;
long b = 2;
while (e != 0) {
if ((e & 1) == 1) {
ans *= b;
ans %= MOD;
}
b *= b;
b %= MOD;
e >>= 1;
}
return ans;
}
public int countWays(int[][] ranges) {
int cnt = 1;
Arrays.sort(ranges, (a, b) -> a[0] != b[0] ? a[0] - b[0] : a[1] - b[1]);
int[] curr = ranges[0];
for (int i = 1; i < ranges.length; i++) {
if (ranges[i][1] < curr[0] || ranges[i][0] > curr[1]) {
++cnt;
curr = ranges[i];
} else {
curr[1] = Math.max(curr[1], ranges[i][1]);
}
}
return (int) powMod(cnt);
}
}