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