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A permutation of the integers 1, 2, ..., n is considered a beautiful arrangement if for every integer i, the following condition is met:

• Either i is divisible by the position it is placed in (i.e., the element at index i is divisible by i), or the position is divisible by the element i (i.e., i is divisible by the element at index i).

Given an integer n, the task is to find how many beautiful arrangements can be constructed.

Clarifying Questions

1. What is the range of the input n?
   • Typically, n will be a small integer, as large values may result in combinatorially large numbers of permutations. The constraints can be found in the problem statement on LeetCode.
2. What is the expected output?
   • The output is a single integer representing the number of beautiful arrangements possible.

Strategy

This problem can be effectively solved using backtracking. We will generate permutations of the integers from 1 to n and for each permutation, we will check if it satisfies the beautiful arrangement condition.

Here is a step-by-step approach:

1. Backtracking Function: Create a function that recursively attempts to place each number in each position.
2. Base Case: If all numbers are placed and the condition holds, count this arrangement as valid.
3. Checking Validity: For each number and position, check if placing the number in the position satisfies the given condition.
4. Recursive Placement: Use a list to keep track of used numbers to avoid duplicates and try placing unplaced numbers in remaining positions.

Code

def countArrangement(n: int) -> int:
    def backtrack(pos):
        # Base case: if pos exceeds n, then we've placed all numbers correctly.
        if pos > n:
            return 1
        
        count = 0
        for num in range(1, n+1):
            if not used[num] and (num % pos == 0 or pos % num == 0):
                used[num] = True
                count += backtrack(pos + 1)
                used[num] = False
        return count
    
    # Initialize the used numbers array
    used = [False] * (n + 1)
    
    # Kick off the backtracking from position 1
    return backtrack(1)

# Example usage:
n = 3
print(countArrangement(n))  # Output: 3

Time Complexity

The time complexity of this solution is O(k), where k depends on the potential number of permutations we process. In the worst case, we might generate all n! permutations. However, due to our constraints and checks, not all permutations are fully generated and many branches are pruned early.

• Space Complexity: O(n) for the used array and recursion call stack.

By implementing and employing backtracking, we efficiently search through possible permutations while pruning invalid configurations early, resulting in an effective solution for relatively small values of n.

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