436. Find Right Interval
Đề Bài
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 * 104intervals[i].length == 2-106 <= starti <= endi <= 106- The start point of each interval is unique.
Thuật Toán & Kỹ Thuật
⏱️ Thời gian
O(log n)
💾 Không gian
O(n)
Lời Giải
Python
0436-find-right-interval.py
class Solution:
def findRightInterval(self, intervals: List[List[int]]) -> List[int]:
s = sorted([(interval[0], i) for i, interval in enumerate(intervals)])
n = len(intervals)
s.append((10**9, -1))
ans = []
for j in range(n):
start, end = intervals[j]
i = bisect_left(s, (end, -1))
# [1, 2], [2, 3], [3, 4]
# 2 1 0
ans.append(s[i][1])
return ans