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

1642. Furthest Building You Can Reach
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);
}