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`