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;
    }
}

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