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[6/21] 1642. Furthest Building You Can Reach

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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

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학부유학생님의 댓글

  • 학부유학생
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Runtime: 866 ms, faster than 44.92% of Python3 online submissions for Furthest Building You Can Reach.
Memory Usage: 28.7 MB, less than 16.28% of Python3 online submissions for Furthest Building You Can Reach.
import collections, heapq
class Solution:
    def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:
        reached = 0
        
        max_heap = [] # bricks_used
        heapq.heapify(max_heap)
        
        for idx, h in enumerate(heights[:-1]):
            if heights[idx+1] <= h: 
                reached+=1
                continue
                
            diff = heights[idx+1] - h
            if bricks < diff:
                if not ladders: break
                
                # ========= Same as Below if statement======
                # if not max_heap or diff > -1*max_heap[0]:
                #     ladders -= 1
                # else:
                #     bricks += -1*heapq.heappop(max_heap)
                #     heapq.heappush(max_heap, -1*diff)
                #     bricks -= diff
                #     ladders -= 1
                # ===============================
                #=============Same as Above if-else
                if max_heap and diff < -1*max_heap[0]:
                    bricks += -1*heapq.heappop(max_heap)
                    heapq.heappush(max_heap, -1*diff)
                    bricks -= diff
                ladders -= 1
                # =================================
            
            else:
                bricks -= diff
                heapq.heappush(max_heap, -1*diff)
            reached += 1
        
        return reached
            
            
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