Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is
bigger than or equal to the end point of the interval i, which can be called that j is on the "right" of i.
For any interval i, you need to store the minimum interval j's index, which means that the interval j has the
minimum start point to build the "right" relationship for interval i. If the interval j doesn't exist, store -1 for
the interval i. Finally, you need output the stored value of each interval as an array.
Note:
You may assume the interval's end point is always bigger than its start point.
You may assume none of these intervals have the same start point.
Example 1:
Input: [ [1,2] ]
Output: [-1]
Explanation: There is only one interval in the collection, so it outputs -1.
Example 2:
Input: [ [3,4], [2,3], [1,2] ]
Output: [-1, 0, 1]
Explanation: There is no satisfied "right" interval for [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point;
For [1,2], the interval [2,3] has minimum-"right" start point.
Example 3:
Input: [ [1,4], [2,3], [3,4] ]
Output: [-1, 2, -1]
Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point.
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method
signature.
public int[] findRightInterval(int[][] intervals) {
int[][] starts = new int[intervals.length][2];
for(int i = 0; i < starts.length; i++)
starts[i] = new int[] {intervals[i][0], i};
Arrays.sort(starts, (a,b)->Integer.compare(a[0],b[0]));
int[] result = new int[intervals.length];
Arrays.fill(result, -1);
for(int i = 0; i < intervals.length; i++) {
int end = intervals[i][1];
int l = 0, r = starts.length-1;
while(l < r) {
int mid = l + (r-l) / 2;
int start = starts[mid][0];
if(start >= end) {
r = mid;
} else {
l = mid + 1;
}
}
if(starts[l][0] >= end && i != starts[l][1]) {
result[i] = starts[l][1];
}
}
return result;
}
}