Class Solution


  • public class Solution
    extends Object
    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
    • Constructor Detail

      • Solution

        public Solution()
    • Method Detail

      • furthestBuilding

        public int furthestBuilding​(int[] heights,
                                    int bricks,
                                    int ladders)