436. Find Right Interval

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.


public class Solution {
    public int[] findRightInterval(Interval[] intervals) {
        int[] result = new int[intervals.length];
        java.util.NavigableMap<Integer, Integer> intervalMap = new TreeMap<>();
        
        for (int i = 0; i < intervals.length; ++i) {
            intervalMap.put(intervals[i].start, i);    
        }
        
        for (int i = 0; i < intervals.length; ++i) {
            Map.Entry<Integer, Integer> entry = intervalMap.ceilingEntry(intervals[i].end);
            result[i] = (entry != null) ? entry.getValue() : -1;
        }
        
        return result;
    }
}

If we are not allowed to use TreeMap:

Sort starts

For each end, find leftmost start using binary search

To get the original index, we need a map

public int[] findRightInterval(Interval[] intervals) {
Map<Integer, Integer> map = new HashMap<>();
List starts = new ArrayList<>();
for (int i = 0; i < intervals.length; i++) {
map.put(intervals[i].start, i);
starts.add(intervals[i].start);
}

 Collections.sort(starts);
 int[] res = new int[intervals.length];
 for (int i = 0; i < intervals.length; i++) {
     int end = intervals[i].end;
     int start = binarySearch(starts, end);
     if (start < end) {
         res[i] = -1;
     } else {
         res[i] = map.get(start);
     }
 }
 return res;
}

public int binarySearch(List list, int x) {
int left = 0, right = list.size() - 1;
while (left < right) {
int mid = left + (right - left) / 2;
if (list.get(mid) < x) {
left = mid + 1;
} else {
right = mid;
}
}
return list.get(left);
}