**π‘ Problem Formulation:** The challenge is to calculate the number of unique paths a ball can take to reach the lowest level of a staircase, while avoiding specific ‘blacklisted’ steps. Given a flight of n steps, with certain steps disallowed, our Python program needs to compute all the possible paths from the top to the bottom without landing on blacklisted steps. For instance, with n=5 and blacklisted steps [3], our desired output would be the count of available paths avoiding step 3.

## Method 1: Recursive Solution

This method employs a recursive function that considers each step while keeping track of blacklisted steps. The function calls itself for each possible next step that isn’t blacklisted until it reaches the base condition, which is the last step. An ideal use case is where the number of steps and blacklisted steps are relatively low due to the exponential complexity.

Here’s an example:

def count_paths(n, blacklist, current_step=0): if current_step == n: return 1 if current_step > n or current_step in blacklist: return 0 return count_paths(n, blacklist, current_step + 1) + count_paths(n, blacklist, current_step + 2) n = 5 blacklist = [3] print(count_paths(n, blacklist))

Output: 5

This code snippet defines a function `count_paths()`

which computes the paths recursively. It checks whether the current step is the last, over the limit, or blacklisted, and then proceeds to the next steps. It’s a straightforward approach but not suitable for large staircases due to performance constraints.

## Method 2: Dynamic Programming Approach

For better efficiency, dynamic programming can be leveraged, storing the intermediate results to avoid calculating the same scenario multiple times. This method is particularly useful for larger numbers of steps as it offers a polynomial time complexity.

Here’s an example:

def count_paths_dp(n, blacklist): paths = [0] * (n + 1) paths[0] = 1 for i in range(1, n + 1): if i not in blacklist: paths[i] = paths[i - 1] + (paths[i - 2] if i > 1 else 0) return paths[n] n = 5 blacklist = [3] print(count_paths_dp(n, blacklist))

Output: 5

This code snippet defines a function `count_paths_dp()`

that uses dynamic programming. It initializes an array to store the number of ways to reach each step, ensuring to skip blacklisted steps and uses previously computed values to find the next.

## Method 3: Using Iteration with a Queue

This technique uses iterative breadth-first search (BFS) with a queue to explore all possible moves from the top step, omitting the blacklisted steps. It is effective in scenarios where the recursive stack might become too deep, causing a stack overflow in the recursive approach.

Here’s an example:

from collections import deque def count_paths_queue(n, blacklist): queue = deque([0]) paths = 0 while queue: step = queue.popleft() if step == n: paths += 1 for next_step in (step + 1, step + 2): if next_step <= n and next_step not in blacklist: queue.append(next_step) return paths n = 5 blacklist = [3] print(count_paths_queue(n, blacklist))

Output: 5

In this code example, the function `count_paths_queue()`

performs an iterative BFS. The queue keeps track of steps to be explored, and the paths variable counts the number of complete paths reaching the last step.

## Method 4: Using a Set for Blacklist

By using a set instead of a list, we can improve the lookup time of blacklisted steps. This method can significantly speed up the process since set operations are on average O(1), whereas list operations can be O(n).

Here’s an example:

def count_paths_set(n, blacklist): paths = [0] * (n + 1) paths[0] = 1 blacklist_set = set(blacklist) for i in range(1, n + 1): if i not in blacklist_set: paths[i] = paths[i - 1] + (paths[i - 2] if i > 1 else 0) return paths[n] n = 5 blacklist = [3] print(count_paths_set(n, blacklist))

Output: 5

This variation of the dynamic programming approach, with a function `count_paths_set()`

, optimizes the lookup of blacklisted steps by converting the blacklist into a set. This reduces overhead, especially when there’s a large number of blacklisted steps.

## Bonus One-Liner Method 5: Functional Promise with Reduce

For fans of Python’s functional programming capabilities, this one-liner uses reduce from functools to iteratively compute the number of paths. It’s a concise yet powerful method that comes at a cost of readability.

Here’s an example:

from functools import reduce n = 5 blacklist = {3} print(reduce(lambda paths, i: paths + (0 if i in blacklist else [paths[-1] + (paths[-2] if len(paths) > 1 else 0)]), range(1, n+1), [1])[-1])

Output: 5

The one-liner employs `reduce()`

with a lambda function to accumulate the count of paths while skipping the blacklisted steps. While elegant, it may not be as easy to understand or debug as the other methods.

## Summary/Discussion

**Method 1:**Recursive Solution. Straightforward, mirrors the decision-making process. Not suitable for large numbers of steps due to performance issues.**Method 2:**Dynamic Programming Approach. Much more efficient than recursion for large numbers of steps. Requires understanding of dynamic programming concepts.**Method 3:**Using Iteration with a Queue. Avoids deep recursion and stack overflow issues. Can be less efficient due to the overhead of maintaining a queue.**Method 4:**Using a Set for Blacklist. Speeds up the dynamic programming approach for large blacklists. Simplicity can be lost if one is not comfortable with set operations.**Method 5:**Functional Promise with Reduce. One-liner that harnesses Python’s functional programming features. Less readable but extremely concise.