436. Find Right Interval
Description
You are given an array of intervals
, where intervals[i] = [starti, endi]
and each starti
is unique.
The right interval for an interval i
is an interval j
such that startj >= endi
and startj
is minimized. Note that i
may equal j
.
Return an array of right interval indices for each interval i
. If no right interval exists for interval i
, then put -1
at index i
.
Example 1:
Input: intervals = [[1,2]] Output: [-1] Explanation: There is only one interval in the collection, so it outputs -1.
Example 2:
Input: intervals = [[3,4],[2,3],[1,2]] Output: [-1,0,1] Explanation: There is no right interval for [3,4]. The right interval for [2,3] is [3,4] since start0 = 3 is the smallest start that is >= end1 = 3. The right interval for [1,2] is [2,3] since start1 = 2 is the smallest start that is >= end2 = 2.
Example 3:
Input: intervals = [[1,4],[2,3],[3,4]] Output: [-1,2,-1] Explanation: There is no right interval for [1,4] and [3,4]. The right interval for [2,3] is [3,4] since start2 = 3 is the smallest start that is >= end1 = 3.
Constraints:
1 <= intervals.length <= 2 * 104
intervals[i].length == 2
-106 <= starti <= endi <= 106
- The start point of each interval is unique.
Solutions
Solution 1: Sorting + Binary Search
We can store the start point and index of each interval into an array arr
, and sort it by the start point. Then we iterate through the interval array, for each interval [_, ed]
, we can use binary search to find the first interval whose start point is greater than or equal to ed
, which is its right-side interval. If found, we store its index into the answer array, otherwise, we store -1
.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of the intervals.
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