436 Find Right Interval

436. Find Right Interval

1. Question

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:

1. You may assume the interval's end point is always bigger than its start point.

2. 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);
        }
    }
}

3. Time & Space Complexity

Binary Search: 时间复杂度O(nlogn), 空间复杂度O(n)