## 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 = 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`

## 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]));
if ladders > 0 {
ladders -= 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 ladders = 1;
let res = 4;
assert_eq!(Solution::furthest_building(heights, bricks, ladders), res);
let heights = vec![4, 12, 2, 7, 3, 18, 20, 3, 19];
let bricks = 10;
let ladders = 2;
let res = 7;
assert_eq!(Solution::furthest_building(heights, bricks, ladders), res);
let heights = vec![14, 3, 19, 3];
let bricks = 17;
let ladders = 0;
let res = 3;
assert_eq!(Solution::furthest_building(heights, bricks, ladders), res);
}
```

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