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

class Solution {
    public int furthestBuilding(int[] heights, int bricks, int ladders) {
        //1.define a heap
        PriorityQueue<Integer> pq = new PriorityQueue<>((x,y)->x-y);
        //2.for loop the heights,
        //2.1 if heap size is bigger than ladders, than we need to poll the least height one to use bricks
        for(int i=0;i<heights.length;i++){
            if(i>0 && heights[i]>heights[i-1]){
                int diff = heights[i]-heights[i-1];
                pq.offer(diff);
                if(pq.size()>ladders){
                    bricks-=pq.poll();
                }
                if(bricks<0) return i-1;
            }
        }
        return heights.length-1;
    }
}