Furthest Building You Can Reach

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`