0436. Find Right Interval
https://leetcode.com/problems/find-right-interval
Description
You are given an array of intervals
, where intervals[i] = [starti, endi]
and each starti
is unique.
The rightinterval for an interval i
is an interval j
such that startj``>= endi
and startj
is minimized.
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.
ac
class Solution {
public int[] findRightInterval(Interval[] intervals) {
// edge cases
if (intervals == null || intervals.length == 0) return new int[0];
int n = intervals.length;
TreeMap<Integer, Integer> tmap = new TreeMap<>();
for (int i = 0; i < n ; i++) {
if (!tmap.containsKey(intervals[i].start)) {
tmap.put(intervals[i].start, i);
}
}
int[] res = new int[n];
for (int i = 0; i < n; i++) {
Integer ceil = tmap.ceilingKey(intervals[i].end);
if (ceil == null) {
res[i] = -1;
} else {
res[i] = tmap.get(ceil);
}
}
return res;
}
}
/*
1) TreeMap record every start val end its first index; 2) iterate, find ceiling of start, get the val
*/
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