# 436. Find Right Interval

Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the "right" of i.

For any interval i, you need to store the minimum interval j's index, which means that the interval j has the minimum start point to build the "right" relationship for interval i. If the interval j doesn't exist, store -1 for the interval i. Finally, you need output the stored value of each interval as an array.

Note: You may assume the interval's end point is always bigger than its start point. You may assume none of these intervals have the same start point. Example 1:

Input: [ [1,2] ]
Output: [-1]

Explanation: There is only one interval in the collection, so it outputs -1. Example 2:

Input: [ [3,4], [2,3], [1,2] ]
Output: [-1, 0, 1]

Explanation: There is no satisfied "right" interval for [3,4]. For [2,3], the interval [3,4] has minimum-"right" start point; For [1,2], the interval [2,3] has minimum-"right" start point. Example 3:

Input: [ [1,4], [2,3], [3,4] ]
Output: [-1, 2, -1]

Explanation: There is no satisfied "right" interval for [1,4] and [3,4]. For [2,3], the interval [3,4] has minimum-"right" start point.

# Solution

Approach 1: brute force. For each interval, iterate through all intervals and remember the index of the nearest right. O(N^2) time.

Approach 2: sort intervals by their start points, then for each interval, take its end point, and use binary search to find it's nearest right. O(NlogN) time.

# Code (Python)

Approach 2:

# Definition for an interval.
# class Interval:
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

import bisect

class Solution:
    def findRightInterval(self, intervals):
        :type intervals: List[Interval]
        :rtype: List[int]
        sorted_indices = sorted(list(range(len(intervals))), key=lambda ind:intervals[ind].start)
        sorted_starting_points = [intervals[ind].start for ind in sorted_indices]
        result = []
        for interval in intervals:
            # assuming end > start for any interval, so bisect_left won't be returning the index of itself
            insert_point = bisect.bisect_left(sorted_starting_points, interval.end)
            if insert_point < len(intervals):
        return result

# Note: 
# sorted(iterable, key) can be implemented using merge sort -- need to turn the iterable into a list first though
# bisect_left(array, target) can be implemented using binary search

# Code (C++)

Approach 1:

Approach 2: