## 1642. 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`

## Rust Solution

``````struct Solution;

use std::cmp::Reverse;
use std::collections::BinaryHeap;

impl Solution {
fn furthest_building(heights: Vec<i32>, mut bricks: i32, mut ladders: i32) -> i32 {
let n = heights.len();
let mut replacement: BinaryHeap<Reverse<i32>> = BinaryHeap::new();
for i in 1..n {
if heights[i - 1] < heights[i] {
replacement.push(Reverse(heights[i] - heights[i - 1]));
} else {
let min = replacement.pop().unwrap().0;
if min <= bricks {
bricks -= min;
} else {
return (i - 1) as i32;
}
}
}
}
(n - 1) as i32
}
}

#[test]
fn test() {
let heights = vec![4, 2, 7, 6, 9, 14, 12];
let bricks = 5;
let res = 4;
let heights = vec![4, 12, 2, 7, 3, 18, 20, 3, 19];
let bricks = 10;