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Create find-right-interval.py
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Python/find-right-interval.py

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# Time: O(nlogn)
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# Space: O(n)
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# Given a set of intervals, for each of the interval i,
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# check if there exists an interval j whose start point is bigger than or
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# equal to the end point of the interval i, which can be called that j is on the "right" of i.
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#
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# For any interval i, you need to store the minimum interval j's index,
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# which means that the interval j has the minimum start point to
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# build the "right" relationship for interval i. If the interval j doesn't exist,
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# store -1 for the interval i. Finally, you need output the stored value of each interval as an array.
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#
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# Note:
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# You may assume the interval's end point is always bigger than its start point.
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# You may assume none of these intervals have the same start point.
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# Example 1:
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# Input: [ [1,2] ]
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#
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# Output: [-1]
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#
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# Explanation: There is only one interval in the collection, so it outputs -1.
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# Example 2:
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# Input: [ [3,4], [2,3], [1,2] ]
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#
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# Output: [-1, 0, 1]
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#
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# Explanation: There is no satisfied "right" interval for [3,4].
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# For [2,3], the interval [3,4] has minimum-"right" start point;
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# For [1,2], the interval [2,3] has minimum-"right" start point.
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# Example 3:
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# Input: [ [1,4], [2,3], [3,4] ]
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#
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# Output: [-1, 2, -1]
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#
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# Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
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# For [2,3], the interval [3,4] has minimum-"right" start point.
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#
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# Definition for an interval.
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# class Interval(object):
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# def __init__(self, s=0, e=0):
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# self.start = s
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# self.end = e
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class Solution(object):
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def findRightInterval(self, intervals):
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"""
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:type intervals: List[Interval]
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:rtype: List[int]
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"""
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sorted_intervals = sorted((interval.start, i) for i, interval in enumerate(intervals))
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result = []
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for interval in intervals:
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idx = bisect.bisect_left(sorted_intervals, (interval.end,))
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result.append(sorted_intervals[idx][1] if idx < len(sorted_intervals) else -1)
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return result

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