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:

- 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.

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