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.
2. Implementation
(1) Binary Search
/**
* Definition for an interval.
* public class Interval {
* int start;
* int end;
* Interval() { start = 0; end = 0; }
* Interval(int s, int e) { start = s; end = e; }
* }
*/
class Solution {
public int[] findRightInterval(Interval[] intervals) {
if (intervals == null || intervals.length == 0) {
return new int[0];
}
int n = intervals.length;
int[] res = new int[n];
List<Integer> startPoints = new ArrayList<>();
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < n; i++) {
startPoints.add(intervals[i].start);
map.put(intervals[i].start, i);
}
Collections.sort(startPoints);
for (int i = 0; i < n; i++) {
int start = binarySearch(startPoints, intervals[i].end);
if (start < intervals[i].end) {
res[i] = -1;
}
else {
res[i] = map.get(start);
}
}
return res;
}
public int binarySearch(List<Integer> list, int target) {
int start = 0, end = list.size() - 1, mid = 0;
while (start + 1 < end) {
mid = start + (end - start) / 2;
if (list.get(mid) < target) {
start = mid + 1;
}
else {
end = mid;
}
}
if (list.get(start) >= target) {
return list.get(start);
}
else {
return list.get(end);
}
}
}