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


#Solution

class Solution {
    public int furthestBuilding(int[] heights, int bricks, int ladders) {
        PriorityQueue<Integer> steps = new PriorityQueue<>(Collections.reverseOrder());
        for(int i=0; i< heights.length-1; i++){
            if(heights[i+1]< heights[i]){continue;}
            bricks -= (heights[i+1]-heights[i]);
            steps.add(heights[i+1]-heights[i]);
            
            if(bricks<0){
                bricks += steps.poll();
                if(ladders>0){
                    ladders--;
                }
                else {
                    return i;
                }
            }
        }
        return heights.length - 1;
    }
}