436 - Find Right Interval

2020/08/27

leetcode

Problem

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.

Solution

class Solution:
    def findRightInterval(self, intervals: List[List[int]]) -> List[int]:
        starts = sorted([[a[0], i] for i, a in enumerate(intervals)]) + [[float('inf'), -1]]
        return [starts[bisect.bisect(starts, [x[1]])][1] for x in intervals]