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Java Solution for LeetCode algorithm problems, continually updating
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:**
**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`