1642. Furthest Building You Can Reach

Problem

You are given an integer array heights representing the heights of buildings, some bricks, and some ladders.

You start your journey from building 0 and move to the next building by possibly using bricks or ladders.

While moving from building i to building i+1 (0-indexed),

• If the current building's height is greater than or equal to the next building's height, you do not need a ladder or bricks.

• If the current building's height is less than the next building's height, you can either use one ladder or (h[i+1] - h[i]) bricks.

Return the furthest building index (0-indexed) you can reach if you use the given ladders and bricks optimally.

  Example 1:

Input: heights = [4,2,7,6,9,14,12], bricks = 5, ladders = 1
Output: 4
Explanation: Starting at building 0, you can follow these steps:
- Go to building 1 without using ladders nor bricks since 4 >= 2.
- Go to building 2 using 5 bricks. You must use either bricks or ladders because 2 < 7.
- Go to building 3 without using ladders nor bricks since 7 >= 6.
- Go to building 4 using your only ladder. You must use either bricks or ladders because 6 < 9.
It is impossible to go beyond building 4 because you do not have any more bricks or ladders.

Example 2:

Input: heights = [4,12,2,7,3,18,20,3,19], bricks = 10, ladders = 2
Output: 7

Example 3:

Input: heights = [14,3,19,3], bricks = 17, ladders = 0
Output: 3

  Constraints:

• 1 <= heights.length <= 105

• 1 <= heights[i] <= 106

• 0 <= bricks <= 109

• 0 <= ladders <= heights.length

Solution

/**
 * @param {number[]} heights
 * @param {number} bricks
 * @param {number} ladders
 * @return {number}
 */
var furthestBuilding = function(heights, bricks, ladders) {
    var queue = new MinPriorityQueue();
    for (var i = 0; i < heights.length - 1; i++) {
        if (heights[i + 1] <= heights[i]) continue;
        queue.enqueue(heights[i + 1] - heights[i], heights[i + 1] - heights[i]);
        if (queue.size() > ladders) {
            var height = queue.dequeue().element;
            if (bricks < height) return i;
            bricks -= height;
        }
    }
    return heights.length - 1;
};

Explain:

nope.

Complexity:

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