What is the time complexity of Beautiful Arrangement?

The optimal solution has a time complexity of O(n!).

What data structures are used to solve Beautiful Arrangement?

This problem uses Array, Dynamic Programming, Backtracking, Bit Manipulation, Bitmask. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Beautiful Arrangement?

Insight 1: Backtracking is suitable because we need to explore all possible permutations of numbers from 1 to n. Insight 2: The small constraint of n <= 15 makes backtracking feasible, as the number of permutations grows factorially, but is still manageable within this limit.

Which companies ask LeetCode 526. Beautiful Arrangement in interviews?

This problem is frequently asked at Cisco, HashedIn, MathWorks, Salesforce, UBS.

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Beautiful Arrangement - LeetCode 526 Solution

Q: How to solve Beautiful Arrangement?

The provided solutions use a backtracking approach to explore all possible permutations. The core idea is to build the arrangement recursively, one position at a time. For each position, we try placing all available numbers (those not yet used). We check if the divisibility condition is satisfied. If it is, we mark the number as used, recursively call the backtracking function to fill the next position, and then unmark the number as used (backtracking step) to explore other possibilities. The base case for the recursion is when we have filled all positions (pos > n), at which point we have found a valid beautiful arrangement, and we increment the count.

Q: Which companies ask LeetCode 526. Beautiful Arrangement in interviews?

This problem is frequently asked at Cisco, HashedIn, MathWorks, Salesforce, UBS.

Beautiful Arrangement - Complete Solution Guide

Beautiful Arrangement is LeetCode problem 526, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 di

Detailed Explanation

The problem asks us to find the number of 'beautiful arrangements' of integers from 1 to `n`. A beautiful arrangement is a permutation of these numbers such that for each position `i` in the permutation, either the number at that position (`perm[i]`) is divisible by `i`, or `i` is divisible by the number at that position. Essentially, we need to count how many permutations satisfy this divisibility condition for all indices from 1 to `n`. The input is a single integer `n` (1 <= n <= 15), and the output is the number of beautiful arrangements.

Solution Approach

The provided solutions use a backtracking approach to explore all possible permutations. The core idea is to build the arrangement recursively, one position at a time. For each position, we try placing all available numbers (those not yet used). We check if the divisibility condition is satisfied. If it is, we mark the number as used, recursively call the backtracking function to fill the next position, and then unmark the number as used (backtracking step) to explore other possibilities. The base case for the recursion is when we have filled all positions (pos > n), at which point we have found a valid beautiful arrangement, and we increment the count.

Step-by-Step Algorithm

Step 1: Initialize a `used` boolean array of size `n+1` to keep track of which numbers have already been placed in the arrangement. Initially, all values are false, meaning no numbers have been used.
Step 2: Define a recursive function `backtrack(pos)` which represents the current position we are trying to fill in the arrangement (ranging from 1 to `n`).
Step 3: Base Case: If `pos` is greater than `n`, it means we have filled all positions from 1 to `n` successfully. In this case, we have found a beautiful arrangement, so we return 1.
Step 4: Iterate through numbers from 1 to `n`. For each number `num`, check if it has not been used yet (`used[num] == false`).
Step 5: If the number is not used, check if the divisibility condition holds: either `num % pos == 0` or `pos % num == 0`.
Step 6: If both conditions (not used and divisibility) are true, mark the number as used (`used[num] = true`).
Step 7: Recursively call `backtrack(pos + 1)` to fill the next position. Add the result of this recursive call to the `count`. This explores all possible arrangements starting with the number 'num' at position 'pos'.
Step 8: After the recursive call returns, unmark the number (`used[num] = false`). This is the backtracking step, allowing us to explore other possibilities for the current position.
Step 9: After iterating through all numbers, return the accumulated `count`, which represents the number of beautiful arrangements that can be formed starting from the current position.
Step 10: The main function `countArrangement(n)` initializes the `used` array and calls the `backtrack(1)` function to start the process from the first position (pos = 1). It then returns the final count.

Key Insights

Insight 1: Backtracking is suitable because we need to explore all possible permutations of numbers from 1 to n.
Insight 2: The small constraint of n <= 15 makes backtracking feasible, as the number of permutations grows factorially, but is still manageable within this limit.
Insight 3: Using a 'used' array (or similar data structure) is crucial to avoid reusing numbers in a permutation and ensure that each arrangement is a valid permutation.

Complexity Analysis

Time Complexity: O(n!)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Backtracking, Bit Manipulation, Bitmask.

Companies

Asked at: Cisco, HashedIn, MathWorks, Salesforce, UBS.

Beautiful Arrangement | LeetCode 526

Medium

Array Dynamic Programming Backtracking Bit Manipulation Bitmask

Time: O(n!)Space: O(n)

Solution Code

Available in:

class Solution {
    public int countArrangement(int n) {
        boolean[] used = new boolean[n + 1];

        return backtrack(1, n, used);
    }

    private int backtrack(int pos, int n, boolean[] used) {
        if (pos > n) {
            return 1;
        }

        int count = 0;
        for (int num = 1; num <= n; num++) {
            if (!used[num] && (num % pos == 0 || pos % num == 0)) {
                used[num] = true;
                count += backtrack(pos + 1, n, used);
                used[num] = false;
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 526: Beautiful Arrangement - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Backtracking Bit Manipulation Bitmask

Companies

CiscoHashedInMathWorksSalesforceUBS

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:526

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Initialize a `used` boolean array of size `n+1` to keep track of which numbers have already been placed in the arrangement. Initially, all values are false, meaning no numbers have been used.
2
Step 2: Define a recursive function `backtrack(pos)` which represents the current position we are trying to fill in the arrangement (ranging from 1 to `n`).
3
Step 3: Base Case: If `pos` is greater than `n`, it means we have filled all positions from 1 to `n` successfully. In this case, we have found a beautiful arrangement, so we return 1.
4
Step 4: Iterate through numbers from 1 to `n`. For each number `num`, check if it has not been used yet (`used[num] == false`).
5
Step 5: If the number is not used, check if the divisibility condition holds: either `num % pos == 0` or `pos % num == 0`.
6
Step 6: If both conditions (not used and divisibility) are true, mark the number as used (`used[num] = true`).
7
Step 7: Recursively call `backtrack(pos + 1)` to fill the next position. Add the result of this recursive call to the `count`. This explores all possible arrangements starting with the number 'num' at position 'pos'.
8
Step 8: After the recursive call returns, unmark the number (`used[num] = false`). This is the backtracking step, allowing us to explore other possibilities for the current position.
9
Step 9: After iterating through all numbers, return the accumulated `count`, which represents the number of beautiful arrangements that can be formed starting from the current position.
10
Step 10: The main function `countArrangement(n)` initializes the `used` array and calls the `backtrack(1)` function to start the process from the first position (pos = 1). It then returns the final count.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n!)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n!) because, in the worst case, we explore all possible permutations. The space complexity is O(n) due to the `used` array and the call stack during recursion.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming with Bitmasking

This approach uses dynamic programming and bitmasking to represent the used numbers. The state `dp[mask][i]` stores the number of beautiful arrangements using the numbers represented by the `mask` and ending with the number `i`. The bitmask `mask` represents which numbers have been used already.

Time: O(n * 2^n)Space: O(n * 2^n)

Trade-offs: While DP with bitmasking improves upon the naive backtracking approach by avoiding redundant calculations and memoizing results, it still has exponential time complexity, although the constant factor is better. In practice, it is not necessarily faster than backtracking for n <= 15 due to the overhead of the DP table. Backtracking is generally preferred due to its simpler implementation.

✨

Key Insights

1Insight 1: Backtracking is suitable because we need to explore all possible permutations of numbers from 1 to n.
2Insight 2: The small constraint of n <= 15 makes backtracking feasible, as the number of permutations grows factorially, but is still manageable within this limit.
3Insight 3: Using a 'used' array (or similar data structure) is crucial to avoid reusing numbers in a permutation and ensure that each arrangement is a valid permutation.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 1. The solution correctly handles this by returning 1, as there is only one arrangement [1] which satisfies the condition.
Edge case 2: Small values of n (e.g., n = 2, 3, 4). The backtracking approach systematically explores all permutations, ensuring correctness for small input sizes.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not marking the numbers as used before the recursive call and unmarking them after. This would lead to incorrect results, as it wouldn't correctly explore all valid permutations.

✗Common mistake 2: Not handling the base case properly. The base case of `pos > n` is crucial to stop the recursion and count a valid arrangement.

✗Common mistake 3: Forgetting to check the divisibility condition (`num % pos == 0 || pos % num == 0`) before making the recursive call. This would result in exploring invalid arrangements.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Task scheduling where certain tasks must be scheduled at specific times or with specific resources, and there are divisibility-related constraints. For example, task 'i' must be scheduled at a time 't' that is either a multiple of 'i' or 'i' is a multiple of 't'.

🏦Real-world scenario 2: Combinatorial optimization problems where finding a valid arrangement under certain constraints (similar to the divisibility condition) is important. For instance, arranging network nodes such that bandwidth allocation satisfies certain dependency rules, where dependency could be expressed in terms of divisibility or related mathematical relations.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 14, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Beautiful Arrangement"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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