README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1600-1699/1642.Furthest%20Building%20You%20Can%20Reach/README_EN.md	1962	Weekly Contest 213 Q3	Greedy Array Heap (Priority Queue)

1642. Furthest Building You Can Reach

中文文档

Description

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

Solutions

Solution 1

Python3

class Solution:
    def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:
        h = []
        for i, a in enumerate(heights[:-1]):
            b = heights[i + 1]
            d = b - a
            if d > 0:
                heappush(h, d)
                if len(h) > ladders:
                    bricks -= heappop(h)
                    if bricks < 0:
                        return i
        return len(heights) - 1

Java

class Solution {
    public int furthestBuilding(int[] heights, int bricks, int ladders) {
        PriorityQueue<Integer> q = new PriorityQueue<>();
        int n = heights.length;
        for (int i = 0; i < n - 1; ++i) {
            int a = heights[i], b = heights[i + 1];
            int d = b - a;
            if (d > 0) {
                q.offer(d);
                if (q.size() > ladders) {
                    bricks -= q.poll();
                    if (bricks < 0) {
                        return i;
                    }
                }
            }
        }
        return n - 1;
    }
}

C++

class Solution {
public:
    int furthestBuilding(vector<int>& heights, int bricks, int ladders) {
        priority_queue<int, vector<int>, greater<int>> q;
        int n = heights.size();
        for (int i = 0; i < n - 1; ++i) {
            int a = heights[i], b = heights[i + 1];
            int d = b - a;
            if (d > 0) {
                q.push(d);
                if (q.size() > ladders) {
                    bricks -= q.top();
                    q.pop();
                    if (bricks < 0) {
                        return i;
                    }
                }
            }
        }
        return n - 1;
    }
};

Go

func furthestBuilding(heights []int, bricks int, ladders int) int {
	q := hp{}
	n := len(heights)
	for i, a := range heights[:n-1] {
		b := heights[i+1]
		d := b - a
		if d > 0 {
			heap.Push(&q, d)
			if q.Len() > ladders {
				bricks -= heap.Pop(&q).(int)
				if bricks < 0 {
					return i
				}
			}
		}
	}
	return n - 1
}

type hp struct{ sort.IntSlice }

func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
	a := h.IntSlice
	v := a[len(a)-1]
	h.IntSlice = a[:len(a)-1]
	return v
}