Leetcode: Find Right Interval

Find Right Interval



Similar Problems:


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:

  1. You may assume the interval’s end point is always bigger than its start point.
  2. 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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.


## Blog link: https://code.dennyzhang.com/find-right-interval
## Basic Ideas: hashmap + binary search
##     Build a hashmap to track the mapping from start point to position index
##     Sort the interval list by the start attribute
##     Use the right attribute to identity the first no smaller values of the start list
##
## Complexity: Time O(n*log(n)), Space O(n)
##
# Definition for an interval.
# class Interval:
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

class Solution:
    def findRightInterval(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: List[int]
        """
        length = len(intervals)
        m = {}
        for i in range(0, length): m[intervals[i].start] = i
        res= [-1]*length
        start_list = list(sorted(m.keys()))
        for i in range(0, length):
            interval = intervals[i]
            target = interval.end
            # binary search
            left, right = 0, length-1
            if target < start_list[0] or target > start_list[-1]:
                continue
            while left <= right:
                mid = left+int((right-left)/2)
                v = start_list[mid]
                if v == target:
                    res[i] = m[start_list[mid]]
                    break
                if v < target:
                    left = mid + 1
                else:
                    right = mid - 1

            # I found, skip
            if res[i] == -1:
                res[i] = m[start_list[left]]
        return res
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