436. Find Right Interval
二分+interval
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.
二分找到end 在sorted starts数组里找位置,第一个大于该end的start
```python3
class Solution:
def findRightInterval(self, intervals: List[List[int]]) -> List[int]:
starts = sorted([(v[0],i) for i,v in enumerate(intervals)])
res = []
for s,e in intervals:
i = bisect.bisect_left(starts,(e,))
res.append(starts[i][1] if i < len(intervals) else -1)
return res
```
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