leetcode 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.

You have bricks and ladders in your hands , When the next building is higher than the current building , You have to go up with bricks or ladders ,
If you choose bricks, you need Height difference brick , If you choose a ladder, you only need one .
Ask the farthest building you can reach , From the subscript 0 Start .

Ideas ：

If the height difference is very large , Need a lot of bricks , But only a ladder is needed ,
An intuitive idea is that those with the greatest height difference use ladders , The rest is made of bricks , To maximize resources , go further .

Obviously, the height difference is not in good order , You can't sort , After all, the building is right there and can't be moved .
So the idea is to use bricks first , When there are not enough bricks, replace the place with a ladder where the height difference is the largest so far , At the same time, recycle the bricks for next use .

So how to find the biggest height difference so far , Use the maximum heap .

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

It says that time efficiency is not high , Because the height difference must be added to at every step queue in

public int furthestBuilding(int[] heights, int bricks, int ladders) {
    
    int n = heights.length;
    if(n == 1) return 0;
    
    
    PriorityQueue<Integer> queue = new PriorityQueue<>(Collections.reverseOrder());
    
    int i;
    for(i = 1; i < heights.length; i++) {
    
        int heightDiff = heights[i] - heights[i - 1];
        
        if(heightDiff <= 0) {
    
            continue;
        } 
        
        if(heightDiff <= bricks) {
    
            bricks -= heightDiff;
            queue.offer(heightDiff);
        } else if(ladders > 0){
    
            if(!queue.isEmpty() && queue.peek() > heightDiff) {
     // When it is necessary to change 
                bricks += queue.poll();
                bricks -= heightDiff;
                queue.offer(heightDiff);
            } 
            ladders--;
        } else {
    
            break;
        }
    }
    
    return i - 1;
}