find_right_interval

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436. Find Right Interval (Medium)

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.

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

Related Topics

[Binary Search]

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