## Problem Statement

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
```

Try here before watching the video.

### Video Solution

### Java Code

```
class Solution {
public int furthestBuilding(int[] heights, int bricks, int ladders) {
PriorityQueue<Integer> diff = new PriorityQueue<>((n1, n2) -> (n2 - n1));
int i;
for(i = 0; i < heights.length - 1; i++) {
if(heights[i] >= heights[i + 1]) continue;
int d = heights[i + 1] - heights[i];
if(bricks >= d) {
bricks -= d;
diff.add(d);
} else if(ladders > 0){
if(diff.size()>0) {
int x = diff.peek();
if(x > d) {
bricks += x;
diff.poll();
diff.add(d);
bricks -= d;
}
}
ladders -= 1;
} else {
break;
}
}
return i;
}
}
```

### C++ Code

```
class Solution {
public:
int furthestBuilding(vector<int>& heights, int bricks, int ladders) {
priority_queue<int> diff;
int i;
for(i = 0; i < heights.size() - 1; i++) {
if(heights[i] >= heights[i + 1]) continue;
int d = heights[i + 1] - heights[i];
if(bricks >= d) {
bricks -= d;
diff.push(d);
} else if(ladders > 0){
if(!diff.empty()) {
int x = diff.top();
if(x > d) {
bricks += x;
diff.pop();
diff.push(d);
bricks -= d;
}
}
ladders -= 1;
} else {
break;
}
}
return i;
}
};
```

### Python Code

```
class Solution(object):
def furthestBuilding(self, heights, bricks, ladders):
diff = []
i = 0
while i < len(heights) - 1:
if heights[i] >= heights[i + 1]:
i += 1
continue
d = heights[i + 1] - heights[i]
if bricks >= d:
bricks -= d
heapq.heappush(diff, -d)
elif ladders > 0:
if diff:
x = -diff[0]
if x > d:
bricks += x
heapq.heappop(diff)
heapq.heappush(diff, -d)
bricks -= d
ladders -= 1
else:
break
i += 1
return i
```

### Javascript Code

```
// remove this comment and paste the code
```

### Go Code

```
// remove this comment and paste the code
```

### Complexity Analysis

**Time Complexity: O(N * logN)Space Complexity: O(N)**