1964. Find the Longest Valid Obstacle Course at Each Position

Problem

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 and i 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 <= 10^5

• 1 <= obstacles[i] <= 10^7

Solution

class Solution {
    public int[] longestObstacleCourseAtEachPosition(int[] obstacles) {
        return longestIncreasingSubsequence(obstacles);
    }

    private int[] longestIncreasingSubsequence(int[] obstacles) {
        int len = 1;
        int length = obstacles.length;
        int[] ans = new int[length];
        int[] arr = new int[length];
        arr[0] = obstacles[0];
        ans[0] = 1;
        for (int i = 1; i < length; i++) {
            int val = obstacles[i];
            if (val >= arr[len - 1]) {
                arr[len++] = val;
                ans[i] = len;
            } else {
                int idx = binarySearch(arr, 0, len - 1, val);
                arr[idx] = val;
                ans[i] = idx + 1;
            }
        }
        return ans;
    }

    private int binarySearch(int[] arr, int lo, int hi, int val) {
        int ans = -1;
        while (lo <= hi) {
            int mid = (lo + hi) / 2;
            if (val >= arr[mid]) {
                lo = mid + 1;
            } else {
                ans = mid;
                hi = mid - 1;
            }
        }
        return ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).