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.
SOLUTION:
import bisect
class Solution:
def findRightInterval(self, intervals: List[List[int]]) -> List[int]:
n = len(intervals)
ans = [-1 for i in range(n)]
starts = []
ends = []
for k in range(n):
i, j = intervals[k]
bisect.insort(starts, (i, k))
bisect.insort(ends, (j, k))
sinx = [s[0] for s in starts]
lst = len(starts)
for end, i in ends:
pos = bisect.bisect_left(sinx, end)
if pos < lst:
ans[i] = starts[pos][1]
return ans
Top comments (0)