# 1642. Furthest Building You Can Reach

#### Medium

***

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:**

![](https://assets.leetcode.com/uploads/2020/10/27/q4.gif)

```
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
```

&#x20;

**Constraints:**

* `1 <= heights.length <= 105`
* `1 <= heights[i] <= 106`
* `0 <= bricks <= 109`
* `0 <= ladders <= heights.length`

```python
class Solution:
    def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:
        arr = []
        for index in range(len(heights)-1):
            # Find Diff Between Two Adj Building
            diff = heights[index+1] - heights[index]
            # If Diff is greater than 0, means we should climb,
            # First we will use ladder
            if diff > 0:
                heapq.heappush(arr, diff)
            # If Ladders are exhausted, then remove bricks equal to shortest height that ladder climbs
            if len(arr) > ladders:
                bricks -= heapq.heappop(arr)
            # if bricks exhausted then this is farthest index
            if bricks < 0:
                return index
        # Default Condition
        return len(heights)-1
```