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g1601_1700.s1642_furthest_building_you_can_reach.readme.md Maven / Gradle / Ivy

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

**Constraints:**

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