1964--find-the-longest-valid-obstacle-course-at-each-position.md

1964. Find the Longest Valid Obstacle Course at Each Position

Difficulty: Hard

https://leetcode.com/problems/find-the-longest-valid-obstacle-course-at-each-position/

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

My Solution(s):

JavaScript

/**
 * @param {number[]} dp
 * @param {number} obstacle
 */
function binarySearch(dp, obstacle) {
    let left = 0;
    let right = dp.length - 1;
    while (left <= right) {
        const mid = (left + right) >>> 1;
        if (dp[mid] <= obstacle) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return left;
}

/**
 * @param {number[]} obstacles
 * @return {number[]}
 */
var longestObstacleCourseAtEachPosition = function (obstacles) {
    const result = [];
    const dp = [];
    const obstaclesLength = obstacles.length;
    for (let i = 0; i < obstaclesLength; i++) {
        const obstacle = obstacles[i];
        let index = binarySearch(dp, obstacle);
        dp[index] = obstacle;
        result.push(index + 1);
    }
    return result;
};

PHP

class Solution
{
    /**
     * @param Integer[] $obstacles
     * @return Integer[]
     */
    function longestObstacleCourseAtEachPosition(array $obstacles): array {
        $dp = [];
        $res = [];
        foreach ($obstacles as $obstacle) {
            $index = $this->binarySearch($dp, $obstacle);
            $dp[$index] = $obstacle;
            $res[] = $index + 1;
        }
        return $res;
    }

    /**
     * @param array $arr
     * @param int $target
     * @return int
     */
    function binarySearch(array $arr, int $target): int {
        $left = 0;
        $right = count($arr) - 1;
        while ($left <= $right) {
            $mid = floor(($left + $right) / 2);
            if ($arr[$mid] <= $target) {
                $left = $mid + 1;
            } else {
                $right = $mid - 1;
            }
        }
        return $left;
    }
}