find-right-interval.py

# Time:  O(nlogn)
# Space: O(n)

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

import bisect


class Solution(object):
    def findRightInterval(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: List[int]
        """
        sorted_intervals = sorted((interval.start, i) for i, interval in enumerate(intervals))
        result = []
        for interval in intervals:
            idx = bisect.bisect_left(sorted_intervals, (interval.end,))
            result.append(sorted_intervals[idx][1] if idx < len(sorted_intervals) else -1)
        return result