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 = 1Output:4Explanation: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 = 2Output:7

**Example 3:**

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

**Constraints:**

`1 <= heights.length <= 10`

^{5}`1 <= heights[i] <= 10`

^{6}`0 <= bricks <= 10`

^{9}`0 <= ladders <= heights.length`

class Solution {
public int furthestBuilding(int[] heights, int bricks, int ladders) {
}
}