**π‘ Problem Formulation:** The task is to determine whether it’s possible to reach a certain position *n* in a sequence, using a set of jumps of varying lengths. If we’re given a starting position (often zero) and a series of potential jumps (represented as an array or list), we must evaluate if reaching the target position *n* is achievable. For instance, given a set of jumps [2, 3, 1] and the target position 5, the desired output is True, as we can jump in sequences like 2 + 3 or 3 + 2 to reach exactly 5.

## Method 1: Recursive Check

This method involves a recursive function `can_reach`

that takes the current position and the array of possible jumps as arguments. It checks if the current position equals *n*, or recursively calls itself with a new position obtained by adding a jump to the current position, thus exploring all possible moves.

Here’s an example:

def can_reach(n, jumps, current=0): if current == n: return True if current > n: return False return any(can_reach(n, jumps, current + jump) for jump in jumps) jumps = [2, 3, 1] print(can_reach(5, jumps))

Output:

True

This snippet defines a function `can_reach()`

that uses recursion to explore all combinations of jumps. If the target is met or exceeded, it returns True or False accordingly. It’s elegant but may become inefficient with large *n* due to the exponential number of recursive calls.

## Method 2: Dynamic Programming

Dynamic programming tackles this problem by breaking it down into subproblems and storing the results, which prevents the re-computation of already-solved subproblems. A boolean list `dp`

is used where each element at index *i* indicates whether position *i* can be reached.

Here’s an example:

def can_reach_dp(n, jumps): dp = [False] * (n + 1) dp[0] = True for i in range(1, n + 1): dp[i] = any(dp[i - jump] for jump in jumps if i - jump >= 0) return dp[n] jumps = [2, 3, 1] print(can_reach_dp(5, jumps))

Output:

True

In this code, a list `dp`

is initialized to represent reachability of each position up to *n*. The algorithm iterates through positions, updating `dp`

based on previous results. This method is more efficient than recursion, particularly for large *n*, but requires additional space.

## Method 3: Iterative Check with Queue

An iterative approach using a queue can also solve the problem, which avoids recursion. This method uses breadth-first search (BFS) to explore positions starting from zero and utilizes a queue to keep track of positions to explore.

Here’s an example:

from collections import deque def can_reach_bfs(n, jumps): visited = set() queue = deque([0]) while queue: position = queue.popleft() if position == n: return True for jump in jumps: new_pos = position + jump if new_pos <= n and new_pos not in visited: visited.add(new_pos) queue.append(new_pos) return False jumps = [2, 3, 1] print(can_reach_bfs(5, jumps))

Output:

True

The function `can_reach_bfs()`

uses a queue to iteratively explore reachable positions. Once a new position is reached, it’s added to a set of visited positions to prevent re-processing. It’s more memory-efficient than the dynamic programming approach and often faster than recursion.

## Method 4: Greedy Approach

Using a greedy algorithm, we constantly choose the largest jump possible at each step. This method assumes that making the largest jump will get us closer to the target. However, this doesn’t always guarantee a solution, making it a heuristic more than a definite method.

Here’s an example:

def can_reach_greedy(n, jumps): position = 0 for jump in sorted(jumps, reverse=True): while position + jump <= n: position += jump return position == n jumps = [2, 3, 1] print(can_reach_greedy(5, jumps))

Output:

True

The function `can_reach_greedy()`

attempts to reach the target position by continually adding the largest jump. This may not work for every set of jumps or every target position, as it might bypass the solution, hence its applicability is limited and may lead to incorrect results.

## Bonus One-Liner Method 5: Using Python Sets

A one-liner approach can utilize the set data structure’s properties in Python to check for reachability. By repeatedly merging sets with updated positions, we can determine if *n* is achievable.

Here’s an example:

can_reach_set = lambda n, jumps: n in set().union(*(range(0, n+1, jump) for jump in jumps)) jumps = [2, 3, 1] print(can_reach_set(5, jumps))

Output:

True

This one-liner uses a generator expression to create ranges for each jump and merges them into one set, checking if the target position *n* is within the result. This is a compact but memory-intensive solution and may not be the most efficient for larger values of *n*.

## Summary/Discussion

**Method 1:**Recursive Check. Intuitive and simple to implement. Inefficient for large input due to the exponential number of calls.**Method 2:**Dynamic Programming. Much more efficient than recursion, particularly with large input sizes. Requires additional memory proportional to*n*.**Method 3:**Iterative Check with Queue. Utilizes a BFS approach. More memory-efficient and can be faster than both recursive and DP methods.**Method 4:**Greedy Approach. Simple and fast for certain inputs. However, it can fail to find a solution even when one exists.**Method 5:**Using Python Sets. This one-liner is compact and easy to understand. Best used for smaller input sizes due to its potential memory usage.