Formatted question description: https://leetcode.ca/all/1964.html

# 1964. Find the Longest Valid Obstacle Course at Each Position

## Level

Hard

## 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 `i-th`

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`

and`i`

**inclusive**. - You must include the
`i-th`

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 <= 10^5`

`1 <= obstacles[i] <= 10^7`

## Solution

For each element in `obstacles`

, find the longest increasing subsequence that ends with the element. Use a greedy approach with binary search to do this.

```
class Solution {
public int[] longestObstacleCourseAtEachPosition(int[] obstacles) {
int length = obstacles.length;
int[] longest = new int[length];
longest[0] = 1;
int len = 1;
int[] d = new int[length + 1];
d[len] = obstacles[0];
for (int i = 1; i < length; ++i) {
if (obstacles[i] >= d[len]) {
d[++len] = obstacles[i];
longest[i] = len;
} else {
int l = 1, r = len, pos = 0;
while (l <= r) {
int mid = (l + r) >> 1;
if (d[mid] <= obstacles[i]) {
pos = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
d[pos + 1] = obstacles[i];
longest[i] = pos + 1;
}
}
return longest;
}
}
```