[99클럽 코테 스터디 30일차 TIL] Find Right Interval

💡문제

https://leetcode.com/problems/find-right-interval/submissions/1362404598/

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.

 

💡키워드

• 이분탐색

 

💡접근/해결 방법

✅ 풀이

import bisect

class Solution:
    def findRightInterval(self, intervals: List[List[int]]) -> List[int]:
        n = len(intervals)
        # 시작점과 그 인덱스를 저장한 리스트를 만듭니다.
        starts = sorted((interval[0], i) for i, interval in enumerate(intervals))
        result = []

        for interval in intervals:
            end = interval[1]
            # 이진 탐색을 통해 끝점보다 크거나 같은 가장 작은 시작점을 찾습니다.
            idx = bisect.bisect_left(starts, (end,))
            if idx < n:
                result.append(starts[idx][1])
            else:
                result.append(-1)

        return result