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The idea is to generate permutations of the numbers from 1 to n and check if each permutation is a beautiful arrangement. This can be efficiently done using backtracking. While generating permutations, we can directly test the divisibility conditions to ensure they form a beautiful arrangement.
Time Complexity: O(k), where k are the valid permutations (worst-case roughly n!).
Space Complexity: O(n) due to the recursion stack.
1public class Solution {
2 public int CountArrangement(int n) {
3 return Count(n, 1, new bool[n + 1]);
4 }
5
6 private int Count(int n, int pos, bool[] visited) {
7 if (pos > n) return 1;
8 int total = 0;
9 for (int i = 1; i <= n; i++) {
10 if (!visited[i] && (i % pos == 0 || pos % i == 0)) {
11 visited[i] = true;
12 total += Count(n, pos + 1, visited);
13 visited[i] = false;
14 }
15 }
16 return total;
17 }
18}The C# solution follows the same backtracking mechanism as other languages explained above, using a boolean array to track visited indices within a recursive function to count each beautiful arrangement.
This approach involves using dynamic programming (DP) with bitmasking to efficiently count beautiful arrangements. By representing the set of visited numbers as a bitmask and using memoization, we can avoid redundant calculations.
Time Complexity: O(n * 2^n) due to bitmask permutations.
Space Complexity: O(2^n) for the DP cache plus recursion stack.
1
This Python solution uses a top-down dynamic programming approach with memoization. The function dfs uses a bitmask to denote which integers have been placed, and iteratively places integers satisfying the beautiful arrangement condition.