Formatted question description: https://leetcode.ca/all/1642.html

# 1642. Furthest Building You Can Reach (Medium)

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

Related Topics:
Binary Search, Heap

## Solution 1. Min heap

Here we only consider positive diffs, and ignore non-positive ones.

We use brick for small diff, and ladder for large diff.

Scan from left to right, maintain a data structure that can efficiently:

1. sort the diffs
2. sum the smallest count - L diffs, where count is the count of (positive) diffs we’ve scanned.

Imagine that we already have the diffs sorted.

[small -----------------------------> large ]
// we can split them in this way.
[ sum of count - L items ][ largest L items ]


So we can use a min heap to keep track of the largest L items.

We keep pushing diff into the heap. When it has more than L items, we push the top element to keep heap of size L and add the popped element to sum. The sum is the count of bricks that we need.

When sum is greater than B, we should stop there.

// OJ: https://leetcode.com/problems/furthest-building-you-can-reach/

// Time: O(NlogL)
// Space: O(L)
class Solution {
public:
int furthestBuilding(vector<int>& H, int B, int L) {
int N = H.size(), sum = 0;
priority_queue<int, vector<int>, greater<>> pq; // min-heap, keep track of the L largest diffs
for (int i = 1; i < N; ++i) {
int diff = H[i] - H[i - 1];
if (diff <= 0) continue;
pq.push(diff);
if (pq.size() <= L) continue;
sum += pq.top();
pq.pop();
if (sum > B) return i - 1;
}
return N - 1;
}
};


Java

class Solution {
public int furthestBuilding(int[] heights, int bricks, int ladders) {
PriorityQueue<Integer> priorityQueue = new PriorityQueue<Integer>(new Comparator<Integer>() {
public int compare(Integer num1, Integer num2) {
return num2 - num1;
}
});
int maxIndex = 0;
int length = heights.length;
for (int i = 1; i < length && bricks >= 0; i++) {
if (heights[i] <= heights[i - 1])
maxIndex = i;
else {
int difference = heights[i] - heights[i - 1];
bricks -= difference;
priorityQueue.offer(difference);
while (bricks < 0 && ladders > 0 && !priorityQueue.isEmpty()) {
bricks += priorityQueue.poll();