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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0501_0600.s0526_beautiful_arrangement;

// #Medium #Array #Dynamic_Programming #Bit_Manipulation #Backtracking #Bitmask
// #2022_07_28_Time_3_ms_(98.66%)_Space_41.5_MB_(24.19%)

/**
 * 526 - Beautiful Arrangement\.
 *
 * Medium
 *
 * Suppose you have `n` integers labeled `1` through `n`. A permutation of those `n` integers `perm` ( **1-indexed** ) is considered a **beautiful arrangement** if for every `i` (`1 <= i <= n`), **either** of the following is true:
 *
 * *   `perm[i]` is divisible by `i`.
 * *   `i` is divisible by `perm[i]`.
 *
 * Given an integer `n`, return _the **number** of the **beautiful arrangements** that you can construct_.
 *
 * **Example 1:**
 *
 * **Input:** n = 2
 *
 * **Output:** 2
 *
 * **Explanation:** 
 *
 * The first beautiful arrangement is [1,2]: 
 *
 * - perm[1] = 1 is divisible by i = 1 
 *
 * - perm[2] = 2 is divisible by i = 2 
 *   
 * The second beautiful arrangement is [2,1]: 
 *
 * - perm[1] = 2 is divisible by i = 1 
 *
 * - i = 2 is divisible by perm[2] = 1
 *
 * **Example 2:**
 *
 * **Input:** n = 1
 *
 * **Output:** 1
 *
 * **Constraints:**
 *
 * *   `1 <= n <= 15`
**/
public class Solution {
    public int countArrangement(int n) {
        return backtrack(n, n, new Integer[1 << (n + 1)], 0);
    }

    private int backtrack(int n, int index, Integer[] cache, int cacheindex) {
        if (index == 0) {
            return 1;
        }
        int result = 0;
        if (cache[cacheindex] != null) {
            return cache[cacheindex];
        }
        for (int i = n; i > 0; i--) {
            if ((cacheindex & (1 << i)) == 0 && (i % (index) == 0 || (index) % i == 0)) {
                result += backtrack(n, index - 1, cache, cacheindex | 1 << i);
            }
        }
        cache[cacheindex] = result;
        return result;
    }
}