💡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
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