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436. Find Right Interval

Array Binary Search Sorting

Problem - Find Right Interval

Medium

You are given an array of intervals, where intervals[i] = [starti, endi] and each starti is unique.

The right interval for an interval i is an interval j such that startj >= endi and startj is minimized. Note that i may equal j.

Return an array of right interval indices for each interval i. If no right interval exists for interval i, then put -1 at index i.

 

Example 1:

Input: intervals = [[1,2]]
Output: [-1]
Explanation: There is only one interval in the collection, so it outputs -1.

Example 2:

Input: intervals = [[3,4],[2,3],[1,2]]
Output: [-1,0,1]
Explanation: There is no right interval for [3,4].
The right interval for [2,3] is [3,4] since start0 = 3 is the smallest start that is >= end1 = 3.
The right interval for [1,2] is [2,3] since start1 = 2 is the smallest start that is >= end2 = 2.

Example 3:

Input: intervals = [[1,4],[2,3],[3,4]]
Output: [-1,2,-1]
Explanation: There is no right interval for [1,4] and [3,4].
The right interval for [2,3] is [3,4] since start2 = 3 is the smallest start that is >= end1 = 3.

 

Constraints:

  • 1 <= intervals.length <= 2 * 104
  • intervals[i].length == 2
  • -106 <= starti <= endi <= 106
  • The start point of each interval is unique.

Solutions

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class Solution:
    def findRightInterval(self, intervals: List[List[int]]) -> List[int]:
        n = len(intervals)
        result = [-1] * n
        temp_arr = sorted((start, i) for i, (start, _) in enumerate(intervals))

        for i, (_, end) in enumerate(intervals):
            j = bisect_left(temp_arr, (end, -inf))
            if j < n:
                result[i] = temp_arr[j][1]
        return result

Submission Stats:

  • Runtime: 35 ms (69.74%)
  • Memory: 21.7 MB (98.40%)