1642. Furthest Building You Can Reach

[#heap]

## description

## 1642. 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),

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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.
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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:

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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:

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Input: heights = [4,12,2,7,3,18,20,3,19], bricks = 10, ladders = 2
Output: 7

Example 3:

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Input: heights = [14,3,19,3], bricks = 17, ladders = 0
Output: 3

–
1 <= heights.length <= 105
–
1 <= heights[i] <= 106
–
0 <= bricks <= 109
–
0 <= ladders <= heights.length

Constraints:

## notes

### Intuition

Ladders can cover any height difference, so they're most valuable for the largest jumps. By greedily using bricks first and retroactively converting the largest brick usage to a ladder when we run out, we maximize our reach.

### Implementation

Use a max heap to track jump sizes. For each positive height difference, assume we use bricks and add the difference to the heap. If bricks go negative, we need a ladder—if none remain, return the current index. Otherwise, use a ladder by popping the largest jump from the heap and adding those bricks back. If we complete the iteration, return the last index.

### Edge-cases

Python's heapq is a min heap, so negate values to simulate a max heap. Remember to negate again when popping.

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Time: O(n log n) — heap operations for each building
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Space: O(n) — heap stores up to n jumps

### Complexity

## solution

python

from typing import List
import heapq

class Solution:
    def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:
        max_heap = []
        for i in range(len(heights) - 1):
            diff = heights[i + 1] - heights[i]
            if diff <= 0:
                continue
            
            bricks -= diff
            heapq.heappush(max_heap, -diff)

            if bricks < 0:
                if ladders == 0:
                    return i
                ladders -= 1
                bricks += -heapq.heappop(max_heap)
        return len(heights) - 1
        



if __name__ == "__main__":
    # Include one-off tests here or debugging logic that can be run by running this file
    # e.g. print(solution.two_sum([1, 2, 3, 4], 3))
    solution = Solution()

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