1964. Find the Longest Valid Obstacle Course at Each Position
Description
You want to build some obstacle courses. You are given a 0-indexed integer array obstacles
of length n
, where obstacles[i]
describes the height of the ith
obstacle.
For every index i
between 0
and n - 1
(inclusive), find the length of the longest obstacle course in obstacles
such that:
- You choose any number of obstacles between
0
andi
inclusive. - You must include the
ith
obstacle in the course. - You must put the chosen obstacles in the same order as they appear in
obstacles
. - Every obstacle (except the first) is taller than or the same height as the obstacle immediately before it.
Return an array ans
of length n
, where ans[i]
is the length of the longest obstacle course for index i
as described above.
Example 1:
Input: obstacles = [1,2,3,2] Output: [1,2,3,3] Explanation: The longest valid obstacle course at each position is: - i = 0: [1], [1] has length 1. - i = 1: [1,2], [1,2] has length 2. - i = 2: [1,2,3], [1,2,3] has length 3. - i = 3: [1,2,3,2], [1,2,2] has length 3.
Example 2:
Input: obstacles = [2,2,1] Output: [1,2,1] Explanation: The longest valid obstacle course at each position is: - i = 0: [2], [2] has length 1. - i = 1: [2,2], [2,2] has length 2. - i = 2: [2,2,1], [1] has length 1.
Example 3:
Input: obstacles = [3,1,5,6,4,2] Output: [1,1,2,3,2,2] Explanation: The longest valid obstacle course at each position is: - i = 0: [3], [3] has length 1. - i = 1: [3,1], [1] has length 1. - i = 2: [3,1,5], [3,5] has length 2. [1,5] is also valid. - i = 3: [3,1,5,6], [3,5,6] has length 3. [1,5,6] is also valid. - i = 4: [3,1,5,6,4], [3,4] has length 2. [1,4] is also valid. - i = 5: [3,1,5,6,4,2], [1,2] has length 2.
Constraints:
n == obstacles.length
1 <= n <= 105
1 <= obstacles[i] <= 107
Solutions
Solution 1: Binary Indexed Tree (Fenwick Tree)
We can use a Binary Indexed Tree to maintain an array of the lengths of the longest increasing subsequences.
Then for each obstacle, we query in the Binary Indexed Tree for the length of the longest increasing subsequence that is less than or equal to the current obstacle, suppose it is $l$. Then the length of the longest increasing subsequence of the current obstacle is $l+1$. We add $l+1$ to the answer array, and update $l+1$ in the Binary Indexed Tree.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the number of obstacles.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 |
|