https://leetcode.com/problems/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
- code maxHeap
class Solution {
public int furthestBuilding(int[] heights, int bricks, int ladders) {
PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
for (int i = 0; i < heights.length - 1; i++){
int diff = heights[i+1] - heights[i];
if (diff <= 0){
continue;
}
pq.offer(diff);
bricks -= diff;
if (bricks < 0 && ladders == 0){
return i;
}
if (bricks < 0){
bricks += pq.poll();
ladders --;
}
}
return heights.length - 1;
}
}
- code minHeap
class Solution:
def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:
hp = []
for i in range(1, len(heights)):
diff = heights[i] - heights[i-1]
if diff <= 0:
continue
heapq.heappush(hp, diff)
if len(hp) <= ladders:
continue
bricks -= heapq.heappop(hp)
if bricks < 0:
return i - 1
return len(heights) - 1