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 <= 105

• 1 <= obstacles[i] <= 107

Solution

/**
 * @param {number[]} obstacles
 * @return {number[]}
 */
var longestObstacleCourseAtEachPosition = function(obstacles) {
    var res = Array(obstacles.length).fill(1);
    var stack = [];
    for (var i = 0; i < obstacles.length; i++) {
        if (!stack.length || obstacles[i] >= stack[stack.length - 1]) {
            res[i] = (stack.length || 0) + 1;
            stack.push(obstacles[i]);
        } else {
            var index = binarySearch(stack, obstacles[i]);
            res[i] = index + 1;
            stack[index] = obstacles[i];
        }
    }
    return res;
};

var binarySearch = function(arr, num) {
    var left = 0;
    var right = arr.length - 1;
    while (left < right) {
        var mid = left + Math.floor((right - left) / 2);
        if (arr[mid] <= num) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return left;
};

Explain:

nope.

Complexity:

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