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1642. Furthest Building You Can Reach

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

  • 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;
    }
}

  • 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;
    }
};

  • 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


  • 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
}

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