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

        


            
```