Problem Description

You are given a positive integer n. Your task is to count how many permutations of the integers from 1 to n satisfy a special "beautiful" property.

A permutation is considered beautiful if for every position i (where 1 ≤ i ≤ n), at least one of these conditions holds:

• The number at position i (denoted as perm[i]) is divisible by i
• The position i is divisible by the number at that position perm[i]

For example, if n = 2:

• The permutation [1, 2] is beautiful because:
  • Position 1: perm[1] = 1, and 1 is divisible by 1 ✓
  • Position 2: perm[2] = 2, and 2 is divisible by 2 ✓
• The permutation [2, 1] is beautiful because:
  • Position 1: perm[1] = 2, and 2 is divisible by 1 ✓
  • Position 2: perm[2] = 1, and 2 is divisible by 1 ✓

Both permutations satisfy the beautiful arrangement criteria, so the answer would be 2.

The solution uses a depth-first search (DFS) approach with backtracking. It first pre-computes which numbers can be placed at each position by checking the divisibility conditions. Then it recursively tries to place valid numbers at each position, marking used numbers with a visited array. When all positions are successfully filled (reaching position n + 1), it counts that as one valid beautiful arrangement. The backtracking ensures all possible valid permutations are explored.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: While we could model this as a graph problem (with positions and numbers as nodes), the problem is fundamentally about arranging numbers in positions with specific constraints, not about traversing a graph structure.

Need to solve for kth smallest/largest?

• No: We're not looking for the kth smallest or largest element. We need to count all valid arrangements.

Involves Linked Lists?

• No: The problem deals with permutations of integers in array positions, not linked list operations.

Does the problem have small constraints?

• Yes: The constraint is 1 ≤ n ≤ 15, which is quite small. With n=15, there are 15! possible permutations (about 1.3 trillion), but the small value of n suggests that an exponential-time algorithm is acceptable.

Brute force / Backtracking?

• Yes: Given the small constraints and the need to explore all valid permutations that satisfy specific conditions, backtracking is the ideal approach. We need to:
  • Try placing each available number at each position
  • Check if the placement satisfies the divisibility conditions
  • Recursively continue to the next position if valid
  • Backtrack when we hit invalid states
  • Count all complete valid arrangements

Conclusion: The flowchart correctly leads us to use a Backtracking approach. This makes sense because we need to systematically explore all possible permutations while pruning invalid branches early (when divisibility conditions aren't met), which is exactly what backtracking excels at.

Intuition

When we first look at this problem, we need to find all permutations where each position satisfies a divisibility relationship with its value. The key insight is that not every number can go in every position - there are constraints based on divisibility.

Let's think about this step by step. For position i, we can only place numbers that either divide i or are divisible by i. For example, at position 2, we can only place numbers like 1 (since 2 is divisible by 1), 2 (since 2 is divisible by 2), 4 (since 4 is divisible by 2), 6, 8, etc.

Since we need to count all valid arrangements, we must explore every possible way to place numbers. This naturally leads us to a recursive approach where:

1. We try to fill positions one by one (from position 1 to n)
2. For each position, we only try numbers that satisfy the divisibility constraint
3. We mark numbers as used to avoid reusing them
4. We recursively fill the next position
5. We backtrack by unmarking the number when we return from recursion

The clever optimization here is pre-computing which numbers are valid for each position. Before starting our recursive search, we can build a mapping that tells us: "for position i, these are all the numbers that could potentially go here." This saves us from repeatedly checking divisibility conditions during the recursion.

For instance, if n = 4:

• Position 1 can have: [1, 2, 3, 4] (everything divides 1)
• Position 2 can have: [1, 2, 4] (numbers that divide 2 or are divisible by 2)
• Position 3 can have: [1, 3] (numbers that divide 3 or are divisible by 3)
• Position 4 can have: [1, 2, 4] (numbers that divide 4 or are divisible by 4)

With this pre-computed information, our backtracking becomes efficient - we only explore valid candidates at each step, dramatically reducing the search space compared to trying all permutations and then checking validity.

Learn more about Dynamic Programming, Backtracking and Bitmask patterns.

Solution Approach

The implementation uses a depth-first search (DFS) with backtracking to systematically explore all valid arrangements. Let's break down the key components:

1. Pre-computation of Valid Matches

First, we build a dictionary match where match[i] contains all numbers that can be placed at position i:

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)

This creates our constraint map. For each position i, we check all numbers from 1 to n and store those that satisfy either j % i == 0 (number divides position) or i % j == 0 (position divides number).

2. Tracking Used Numbers

We maintain a boolean array vis of size n + 1 to track which numbers have been used:

vis = [False] * (n + 1)

We use 1-indexed arrays to match the problem's 1-indexed positions, so vis[0] is unused.

3. DFS with Backtracking

The core recursive function dfs(i) tries to fill position i:

def dfs(i):
    nonlocal ans, n
    if i == n + 1:  # Base case: all positions filled
        ans += 1
        return
  
    for j in match[i]:  # Try each valid number for position i
        if not vis[j]:  # If number j hasn't been used
            vis[j] = True      # Mark as used
            dfs(i + 1)         # Recursively fill next position
            vis[j] = False     # Backtrack: unmark for other branches

4. Algorithm Flow

• Start with dfs(1) to fill position 1
• For each position i, iterate through all valid numbers in match[i]
• If a number hasn't been used (vis[j] == False), mark it as used and recursively try to fill the next position
• When we successfully fill all n positions (reach position n + 1), increment our answer counter
• After exploring a branch, backtrack by unmarking the number (vis[j] = False) so it can be used in other arrangements
• The process continues until all valid arrangements are counted

Time Complexity: O(n!) in the worst case, as we might need to explore all permutations. However, the pre-computed constraints significantly prune the search space.

Space Complexity: O(n²) for the match dictionary plus O(n) for the recursion stack and visited array.

Example Walkthrough

Let's walk through the solution with n = 3 to see how the algorithm finds all beautiful arrangements.

Step 1: Pre-compute valid matches

For each position, we determine which numbers can be placed there:

• Position 1: Can place [1, 2, 3] (all numbers divide 1)
• Position 2: Can place [1, 2] (1 divides into 2, and 2 divides 2)
• Position 3: Can place [1, 3] (1 divides into 3, and 3 divides 3)

Step 2: DFS exploration starting from position 1

Initial state: vis = [False, False, False, False], ans = 0

dfs(1) - Fill position 1:
  Try number 1: vis[1] = True
    └─ dfs(2) - Fill position 2:
        Try number 2: vis[2] = True
          └─ dfs(3) - Fill position 3:
              Try number 3: vis[3] = True
                └─ dfs(4) → ans = 1 (Found: [1,2,3])
              vis[3] = False (backtrack)
        vis[2] = False (backtrack)
    vis[1] = False (backtrack)
  
  Try number 2: vis[2] = True
    └─ dfs(2) - Fill position 2:
        Try number 1: vis[1] = True
          └─ dfs(3) - Fill position 3:
              Try number 3: vis[3] = True
                └─ dfs(4) → ans = 2 (Found: [2,1,3])
              vis[3] = False (backtrack)
        vis[1] = False (backtrack)
    vis[2] = False (backtrack)
  
  Try number 3: vis[3] = True
    └─ dfs(2) - Fill position 2:
        Try number 1: vis[1] = True
          └─ dfs(3) - Fill position 3:
              Number 1 already used, skip
              Number 3 already used, skip
              (Dead end - no valid arrangement)
        vis[1] = False (backtrack)
        Try number 2: vis[2] = True
          └─ dfs(3) - Fill position 3:
              Try number 1: vis[1] = True
                └─ dfs(4) → ans = 3 (Found: [3,2,1])
              vis[1] = False (backtrack)
        vis[2] = False (backtrack)
    vis[3] = False (backtrack)

Step 3: Result

The algorithm found 3 beautiful arrangements:

1. [1, 2, 3] - pos1: 1|1✓, pos2: 2|2✓, pos3: 3|3✓
2. [2, 1, 3] - pos1: 2|1✓, pos2: 2|1✓, pos3: 3|3✓
3. [3, 2, 1] - pos1: 3|1✓, pos2: 2|2✓, pos3: 3|1✓

The backtracking ensures we explore all valid paths while the pre-computed matches prevent us from trying invalid placements, making the search efficient.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n!)

The algorithm uses backtracking to explore all valid permutations. In the worst case, we need to explore all possible arrangements. At position 1, we have at most n choices, at position 2 we have at most n-1 choices, and so on. This gives us an upper bound of n! permutations to explore. While the divisibility constraints reduce the actual number of valid paths explored (pruning occurs when match[i] has fewer options or when elements are already visited), the worst-case time complexity remains O(n!).

Space Complexity: O(n²)

The space complexity consists of several components:

• vis array: O(n) space for tracking visited elements
• match dictionary: O(n²) space in the worst case, as each of the n positions could potentially match with all n numbers
• Recursion call stack: O(n) depth in the worst case
• Other variables (ans, i, j): O(1) space

The dominant factor is the match dictionary which stores the precomputed valid matches, resulting in an overall space complexity of O(n²).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Index Handling (0-indexed vs 1-indexed)

One of the most common mistakes is mixing up 0-indexed and 1-indexed arrays. The problem uses 1-indexed positions (positions 1 to n), but Python arrays are naturally 0-indexed.

Incorrect Implementation:

# Wrong: Using 0-indexed positions with 1-indexed problem logic
used = [False] * n  # Creates array of size n
for position in range(n):  # Iterates 0 to n-1
    for number in range(n):  # Iterates 0 to n-1
        if number % position == 0:  # Division by zero when position = 0!
            valid_numbers[position].append(number)

Correct Implementation:

# Correct: Properly handle 1-indexed positions
used = [False] * (n + 1)  # Create array of size n+1, index 0 unused
for position in range(1, n + 1):  # Iterate 1 to n
    for number in range(1, n + 1):  # Iterate 1 to n
        if number % position == 0 or position % number == 0:
            valid_numbers[position].append(number)

2. Forgetting to Backtrack (Not Resetting State)

Failing to reset the used array after exploring a branch will cause incorrect results, as numbers will remain marked as used for subsequent branches.

Incorrect Implementation:

def backtrack(position):
    if position == n + 1:
        count += 1
        return
  
    for number in valid_numbers[position]:
        if not used[number]:
            used[number] = True
            backtrack(position + 1)
            # Missing: used[number] = False  <-- Forgot to backtrack!

Correct Implementation:

def backtrack(position):
    if position == n + 1:
        count += 1
        return
  
    for number in valid_numbers[position]:
        if not used[number]:
            used[number] = True
            backtrack(position + 1)
            used[number] = False  # Essential: Reset state for other branches

3. Using Global Variables Instead of nonlocal

When using nested functions in Python, forgetting to declare nonlocal for variables you want to modify can lead to UnboundLocalError or create new local variables instead of modifying the outer scope variable.

Incorrect Implementation:

def countArrangement(self, n: int) -> int:
    count = 0
  
    def backtrack(position):
        if position == n + 1:
            count += 1  # UnboundLocalError: local variable 'count' referenced before assignment
            return

Correct Implementation:

def countArrangement(self, n: int) -> int:
    count = 0
  
    def backtrack(position):
        nonlocal count  # Declare that we're modifying the outer scope variable
        if position == n + 1:
            count += 1
            return

4. Inefficient Pre-computation or No Pre-computation

Computing valid numbers on-the-fly during DFS instead of pre-computing them once leads to redundant calculations and slower performance.

Inefficient Implementation:

def backtrack(position):
    if position == n + 1:
        count += 1
        return
  
    # Computing valid numbers every time - inefficient!
    for number in range(1, n + 1):
        if not used[number]:
            if number % position == 0 or position % number == 0:
                used[number] = True
                backtrack(position + 1)
                used[number] = False

Efficient Implementation:

# Pre-compute once before DFS
valid_numbers = defaultdict(list)
for position in range(1, n + 1):
    for number in range(1, n + 1):
        if number % position == 0 or position % number == 0:
            valid_numbers[position].append(number)

# Then use pre-computed values in DFS
def backtrack(position):
    if position == n + 1:
        count += 1
        return
  
    for number in valid_numbers[position]:  # Use pre-computed list
        if not used[number]:
            used[number] = True
            backtrack(position + 1)
            used[number] = False