For each $end_i$, find the first $start_j$ that $start_j > end_i$.
Apparently, if we use binary search, we can finish in O(logn).
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.
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.
1 2 3 4 5
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.