Lecture 120:-  Beautiful Arrangement

The "Beautiful Arrangement" problem is a combinatorial problem that involves arranging numbers from 1 to N in such a way that specific conditions are met. In the problem, you are given an integer N, and you need to find the count of all possible beautiful arrangements where:

1. The number at the ith position is divisible by i.
2. i is divisible by the number at the ith position.

Here's how you can approach the "Beautiful Arrangement" problem using a recursive backtracking approach in Python:

pythonCopy code

def count_arrangement(N): def backtrack(start): nonlocal count if start > N: count += 1 return for num in range(1, N + 1): if not visited[num] and (start % num == 0 or num % start == 0): visited[num] = True backtrack(start + 1) visited[num] = False count = 0 visited = [False] * (N + 1) backtrack(1) return count # Example usage N = 3 result = count_arrangement(N) print("Number of beautiful arrangements:", result)

In this example, the count_arrangement function takes an integer N as input and returns the count of all possible beautiful arrangements.

The backtrack function generates beautiful arrangements by trying different numbers at each position. It uses a visited list to keep track of which numbers have already been used.

For each position, the backtrack function iterates through the available numbers and checks whether the number satisfies the conditions for a beautiful arrangement. If the conditions are met, the number is marked as visited, and the function recurses to the next position. After recursion, the number is marked as not visited to backtrack and explore other possibilities.

The time complexity of this solution can vary based on N and the number of valid beautiful arrangements. In the worst case, it can be exponential due to the branching factor of the recursion. However, the use of backtracking helps avoid unnecessary recursion and reduces the number of recursive calls.

This solution generates beautiful arrangements that satisfy the given conditions and counts them. It's important to note that for larger values of N, the problem becomes computationally more challenging due to the exponential nature of the solution.