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Abhishek Chaudhary

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# Find Right 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.

SOLUTION:

``````import bisect

class Solution:
def findRightInterval(self, intervals: List[List[int]]) -> List[int]:
n = len(intervals)
ans = [-1 for i in range(n)]
starts = []
ends = []
for k in range(n):
i, j = intervals[k]
bisect.insort(starts, (i, k))
bisect.insort(ends, (j, k))
sinx = [s[0] for s in starts]
lst = len(starts)
for end, i in ends:
pos = bisect.bisect_left(sinx, end)
if pos < lst:
ans[i] = starts[pos][1]
return ans
``````