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 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 <= 105
• 1 <= obstacles[i] <= 107

 

Solutions

Solution: Binary Search

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

 

JavaScript

/**
 * @param {number[]} obstacles
 * @return {number[]}
 */
const longestObstacleCourseAtEachPosition = function (obstacles) {
  const tails = [];
  const result = [];

  const findFirstGreater = (nums, target) => {
    let left = 0;
    let right = nums.length - 1;

    while (left <= right) {
      const mid = Math.floor((left + right) / 2);

      nums[mid] > target ? (right = mid - 1) : (left = mid + 1);
    }

    return left;
  };

  for (const obstacle of obstacles) {
    if (tails.length && obstacle < tails.at(-1)) {
      const index = findFirstGreater(tails, obstacle);

      tails[index] = obstacle;
      result.push(index + 1);
    } else {
      tails.push(obstacle);
      result.push(tails.length);
    }
  }

  return result;
};