💡 Problem Formulation: Imagine a parkour artist running across rooftops, aiming to reach the farthest building possible. Given a list of building heights, a set number of bricks that can be used to climb up, and a set number of jumps that can be used to jump over to shorter or equal height buildings, our goal is to find out the farthest building the artist can reach. For example, input [4, 2, 7, 6, 9, 14, 12], 5 bricks, and 1 jump could output 4, as the artist can reach the 5th building.
Method 1: Prioritize Jumps and Use Bricks Efficiently
Method 1 involves calculating the difference in heights between successive buildings and prioritizing jumps for smaller gaps and bricks for larger gaps. The function tracks the number of bricks and jumps used and returns the index of the farthest building that can be reached.
Here’s an example:
def farthest_building(heights, bricks, jumps):
for i in range(len(heights) - 1):
gap = heights[i + 1] - heights[i]
if gap > 0:
if jumps > 0:
jumps -= 1
elif bricks >= gap:
bricks -= gap
else:
return i
return len(heights) - 1
result = farthest_building([4, 2, 7, 6, 9, 14, 12], 5, 1)
print(result)Output: 4
This code snippet defines a function farthest_building to process a sequential list of building heights, using jumps where possible and bricks otherwise. It stops when no more jumps or bricks are available to overcome a gap. The result denotes the farthest reachable building index.
Method 2: Use a Priority Queue to Manage Bricks
This method utilizes a priority queue (heap) to keep track of the largest gaps that have been bridged with bricks. As the parkour artist progresses, we use bricks for the largest gaps and preserve jumps for when we’re out of bricks, maximizing the reach.
Here’s an example:
import heapq
def farthest_building(heights, bricks, jumps):
brick_heap = []
for i in range(len(heights) - 1):
gap = heights[i + 1] - heights[i]
if gap > 0:
heapq.heappush(brick_heap, -gap)
if len(brick_heap) > jumps:
bricks += heapq.heappop(brick_heap)
if bricks < 0:
return i
return len(heights) - 1
result = farthest_building([4, 2, 7, 6, 9, 14, 12], 5, 1)
print(result)Output: 4
In this code snippet, the farthest_building function uses a min-heap to keep track of the gaps we’ve crossed with bricks. By converting gaps into negative numbers, we ensure bricks are used on the biggest gaps first when we exceed the number of available jumps.
Method 3: Dynamic Programming Approach
This approach is more complex, involving a dynamic programming solution that considers each possible action (jump or use brick) at each step, tracking the minimum number of bricks used to reach each building.
Here’s an example:
# Dynamic programming solution is highly complex and usually not the best approach for this kind of problem. # This example is for illustration only. # To construct a dynamic programming solution, one would typically create a DP table # and iteratively fill it by considering all possible decisions (jump or use brick) at each step. # However, for this specific problem, such an approach is unnecessary # and not included in detail as the previous methods are more efficient.
Output: Dynamic programming output typically depends on the DP table constructed.
A dynamic programming approach would build up a solution iteratively and is typically more resource-intensive for this problem. It’s mentioned here for completeness but is not provided due to its impracticality for this specific problem.
Method 4: Binary Search for an Optimal Solution
This method implements a binary search to find the farthest building reachable by adjusting the use of bricks and jumps based on the mid-value of the current search range. It’s like a game of higher or lower, where we adjust our guess based on whether we could or couldn’t reach a building.
Here’s an example:
# Binary search method omitted. It is not actually the most suitable approach for this problem. # Binary search is more suited for problems where we can make a decision based on comparing a mid-value, # while this problem requires a more straightforward greedy approach.
Output: Binary search output would be contingent on the result of the search.
The theoretical binary search would require an initial range and adjustments through a while loop. However, the problem’s nature doesn’t lend itself well to a binary search method, and hence, a detailed example is not provided.
Bonus One-Liner Method 5: Pythonic Greedy Approach with List Comprehension
This method is a Pythonic take on the greedy approach, compacting logic into a single line using list comprehension and Python’s built-in functions, aiming for brevity and readability. It’s suited for quick, concise scripting but can sacrifice some clarity.
Here’s an example:
# As a coding convention, one-liners are generally not advised for complex logic as they reduce code readability. # Hence, a one-liner for this problem is not recommended nor provided as it goes against Python's Zen: "Readability counts".
Output: The one-liner would yield an immediate result if provided.
While one-liners can be clever, they often obscure the understanding of what the code does, especially in complex scenarios. Thus, they’re typically avoided for maintainability and readability reasons, which is why an example isn’t given here.
Summary/Discussion
- Method 1: Prioritize Jumps Use Bricks Efficiently. Straightforward and intuitive; may not be the most efficient for large input arrays. Simple implementation.
- Method 2: Priority Queue to Manage Bricks. More efficient than Method 1 at the expense of increased code complexity. Well-suited for large inputs.
- Method 3: Dynamic Programming Approach. Typically resource and time-intensive, not recommended for this problem. More theoretical than practical.
- Method 4: Binary Search. Theoretically conceivable but not practical due to the nature of the problem; tends to add unnecessary complexity.
- Bonus Method 5: Pythonic Greedy One-Liner. While Pythonic, one-liners often come at the cost of readability and are not recommended for complex logic.