LeetCode #526 — MEDIUM

Beautiful Arrangement

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Backtracking · Bit Manipulation

Example 1

Example 2

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #526: Beautiful Arrangement
class Solution {
    private int n;
    private int ans;
    private boolean[] vis;
    private Map<Integer, List<Integer>> match;

    public int countArrangement(int n) {
        this.n = n;
        ans = 0;
        vis = new boolean[n + 1];
        match = new HashMap<>();
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (i % j == 0 || j % i == 0) {
                    match.computeIfAbsent(i, k -> new ArrayList<>()).add(j);
                }
            }
        }
        dfs(1);
        return ans;
    }

    private void dfs(int i) {
        if (i == n + 1) {
            ++ans;
            return;
        }
        if (!match.containsKey(i)) {
            return;
        }
        for (int j : match.get(i)) {
            if (!vis[j]) {
                vis[j] = true;
                dfs(i + 1);
                vis[j] = false;
            }
        }
    }
}

// Accepted solution for LeetCode #526: Beautiful Arrangement
func countArrangement(n int) int {
	ans := 0
	match := make(map[int][]int)
	for i := 1; i <= n; i++ {
		for j := 1; j <= n; j++ {
			if i%j == 0 || j%i == 0 {
				match[i] = append(match[i], j)
			}
		}
	}
	vis := make([]bool, n+1)

	var dfs func(i int)
	dfs = func(i int) {
		if i == n+1 {
			ans++
			return
		}
		for _, j := range match[i] {
			if !vis[j] {
				vis[j] = true
				dfs(i + 1)
				vis[j] = false
			}
		}
	}

	dfs(1)
	return ans
}

# Accepted solution for LeetCode #526: Beautiful Arrangement
class Solution:
    def countArrangement(self, n: int) -> int:
        def dfs(i):
            nonlocal ans, n
            if i == n + 1:
                ans += 1
                return
            for j in match[i]:
                if not vis[j]:
                    vis[j] = True
                    dfs(i + 1)
                    vis[j] = False

        ans = 0
        vis = [False] * (n + 1)
        match = defaultdict(list)
        for i in range(1, n + 1):
            for j in range(1, n + 1):
                if j % i == 0 or i % j == 0:
                    match[i].append(j)

        dfs(1)
        return ans

// Accepted solution for LeetCode #526: Beautiful Arrangement
impl Solution {
    fn dfs(i: usize, n: usize, mat: &Vec<Vec<usize>>, vis: &mut Vec<bool>, res: &mut i32) {
        if i == n + 1 {
            *res += 1;
            return;
        }
        for &j in mat[i].iter() {
            if !vis[j] {
                vis[j] = true;
                Self::dfs(i + 1, n, mat, vis, res);
                vis[j] = false;
            }
        }
    }

    pub fn count_arrangement(n: i32) -> i32 {
        let n = n as usize;
        let mut vis = vec![false; n + 1];
        let mut mat = vec![Vec::new(); n + 1];
        for i in 1..=n {
            for j in 1..=n {
                if i % j == 0 || j % i == 0 {
                    mat[i].push(j);
                }
            }
        }

        let mut res = 0;
        Self::dfs(1, n, &mat, &mut vis, &mut res);
        res
    }
}

// Accepted solution for LeetCode #526: Beautiful Arrangement
function countArrangement(n: number): number {
    const vis = new Array(n + 1).fill(0);
    const match = Array.from({ length: n + 1 }, () => []);
    for (let i = 1; i <= n; i++) {
        for (let j = 1; j <= n; j++) {
            if (i % j === 0 || j % i === 0) {
                match[i].push(j);
            }
        }
    }

    let res = 0;
    const dfs = (i: number, n: number) => {
        if (i === n + 1) {
            res++;
            return;
        }
        for (const j of match[i]) {
            if (!vis[j]) {
                vis[j] = true;
                dfs(i + 1, n);
                vis[j] = false;
            }
        }
    };
    dfs(1, n);
    return res;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.