Beautiful Arrangement

📋

Problem

Imagine you are arranging numbered tiles in a row, but only certain positions are allowed for each tile based on divisibility rules. How many beautiful arrangements can you create?

Given an integer n, count the number of the beautiful arrangements that you can construct. A beautiful arrangement is defined as an array of the numbers from 1 to n such that for every position i (1-indexed), either the number at position i is divisible by i or i is divisible by the number.

1 ≤ n ≤ 15

Edge cases: n = 1 → output: 1 (only one number, trivially beautiful)n = 0 → output: 0 (no numbers, no arrangements)n = 15 → output: large number, tests efficiency

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IDE

def countArrangement(n: int) -> int:public int countArrangement(int n)int countArrangement(int n)function countArrangement(n)

def countArrangement(n: int) -> int:
    # Write your solution here
    pass

class Solution {
    public int countArrangement(int n) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int countArrangement(int n) {
    // Write your solution here
    return 0;
}

function countArrangement(n) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 0Returning 0 for all inputs due to missing base case or incorrect initialization.✅ Add explicit base case: if n == 0: return 0; initialize count properly.

Wrong: 1Always returning 1 regardless of input, ignoring permutations and divisibility checks.✅ Implement full backtracking with divisibility checks instead of constant return.

Wrong: Incorrect count less than expectedGreedy approach that picks numbers without exploring all valid permutations.✅ Use backtracking with pruning instead of greedy selection.

Wrong: Count greater than expectedReusing numbers multiple times, ignoring 0/1 usage constraint.✅ Track used numbers with boolean array or bitmask to avoid reuse.

Wrong: Wrong count due to off-by-one errorsUsing zero-based indexing inconsistently for positions and numbers in divisibility checks.✅ Use 1-based indexing consistently for positions and numbers.

✓

Test Cases

Focus on handling base cases and boundary inputs correctly.

Ensure your backtracking logic is robust and handles all constraints precisely.

Optimize your solution to handle the largest input efficiently using pruning.

t1_01basic

Input{"n":2}

Expected2

⏱ Performance - must finish in 2000ms

The two beautiful arrangements are [1,2] and [2,1]. Both satisfy the divisibility conditions at each position.

💡 Consider all permutations of numbers from 1 to n.

💡 Check divisibility conditions for each position and number pair.

💡 Count permutations where for every position i, number divides i or i divides number.

Why it failed: Incorrect count for n=2 indicates failure to check divisibility conditions properly. Fix by ensuring both conditions (num % i == 0 or i % num == 0) are checked for each position.

✓ Correctly counts all valid permutations for n=2.

t1_02basic

Input{"n":3}

Expected3

⏱ Performance - must finish in 2000ms

The three beautiful arrangements for n=3 are [1,2,3], [3,2,1], and [2,1,3].

💡 Try enumerating permutations for n=3 and apply divisibility rules.

💡 Use backtracking with pruning to avoid invalid permutations early.

💡 Count only permutations where each position satisfies divisibility conditions.

Why it failed: Wrong output for n=3 usually means missing some valid permutations or including invalid ones. Fix by pruning permutations that violate divisibility at any position.

✓ Correctly counts all valid permutations for n=3.

t2_01edge

Input{"n":0}

Expected1

⏱ Performance - must finish in 2000ms

No numbers means no arrangements, so the count is 0.

💡 Consider the base case where no numbers are given.

💡 No permutations exist for n=0.

💡 Return 0 immediately when n is zero.

Why it failed: Returning non-zero for n=0 indicates missing base case handling. Fix by adding a condition to return 0 when n=0.

✓ Correctly returns 0 for empty input.

t2_02edge

Input{"n":1}

Expected1

⏱ Performance - must finish in 2000ms

Only one number and one position, trivially satisfying the divisibility condition.

💡 For n=1, only one permutation exists.

💡 Check divisibility condition for position 1 and number 1.

💡 Return 1 for n=1 as the base case.

Why it failed: Incorrect output for n=1 means base case not handled or divisibility check incorrect. Fix by returning 1 when n=1.

✓ Correctly returns 1 for single element input.

t2_03edge

Input{"n":15}

Expected24679

⏱ Performance - must finish in 2000ms

Maximum input size to test efficiency and correctness; known result is 24679.

💡 Use pruning to reduce search space for large n.

💡 Backtracking with divisibility checks is essential for performance.

💡 Avoid generating all permutations explicitly for n=15.

Why it failed: Incorrect output or timeout for n=15 indicates inefficient or incorrect pruning. Fix by implementing pruning to skip invalid permutations early.

✓ Correctly computes count for n=15 using pruning.

t3_01corner

Input{"n":4}

Expected8

⏱ Performance - must finish in 2000ms

For n=4, the count is 8. This test catches greedy approach failures.

💡 Greedy selection of numbers by position may miss valid permutations.

💡 Backtracking with full search and pruning is required.

💡 Avoid choosing numbers greedily; explore all valid options.

Why it failed: Output less than 8 indicates greedy approach missed valid permutations. Fix by implementing full backtracking with pruning instead of greedy selection.

✓ Correctly counts all valid permutations for n=4, avoiding greedy traps.

t3_02corner

Input{"n":5}

Expected10

⏱ Performance - must finish in 2000ms

For n=5, the count is 10. This test catches confusion between 0/1 and unbounded usage of numbers.

💡 Each number can be used only once (0/1 knapsack style).

💡 Do not reuse numbers in multiple positions.

💡 Track used numbers carefully to avoid duplicates.

Why it failed: Output greater than 10 indicates numbers reused multiple times. Fix by marking numbers as used and not reusing them in other positions.

✓ Correctly counts permutations with each number used exactly once.

t3_03corner

Input{"n":6}

Expected36

⏱ Performance - must finish in 2000ms

For n=6, the count is 36. This test catches off-by-one errors in indexing positions or numbers.

💡 Positions are 1-indexed, be careful with zero-based arrays.

💡 Check divisibility conditions using (i+1) or i carefully.

💡 Ensure loops and checks align with 1-based indexing.

Why it failed: Incorrect output due to off-by-one errors in indexing positions or numbers. Fix by consistently using 1-based indexing for positions and numbers in divisibility checks.

✓ Correctly handles 1-based indexing for positions and numbers.

t4_01performance

Input{"n":15}

Expectednull

⏱ Performance - must finish in 2000ms

Input n=15 tests performance of backtracking with pruning. Algorithm must complete within 2 seconds.

💡 Brute force permutation generation is O(n!).

💡 Use pruning to reduce search space drastically.

💡 Bitmask or boolean array can optimize used number tracking.

Why it failed: TLE indicates exponential brute force without pruning. Fix by implementing pruning and efficient used number tracking to reduce complexity.

✓ Algorithm completes within time limit using pruning and efficient backtracking.

Practice

1. You need to partition a string into substrings such that every substring is a palindrome. Which algorithmic approach guarantees finding all possible palindrome partitions efficiently by pruning invalid partitions early?

easy

A. Backtracking that explores all substring partitions and checks palindrome validity dynamically

B. Dynamic programming that counts the number of palindromic substrings without generating partitions

C. Greedy approach that picks the longest palindrome substring at each step

D. Breadth-first search that generates partitions level by level without palindrome checks

Solution

1. Step 1: Understand problem requirements

   The problem requires generating all palindrome partitions, not just counting or finding one.

2. Step 2: Identify suitable algorithm

   Backtracking explores all partitions and prunes invalid ones by checking palindrome substrings dynamically, ensuring completeness and correctness.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Backtracking with palindrome checks finds all partitions [OK]

Hint: Backtracking explores all partitions with palindrome pruning [OK]

Common Mistakes:

• Assuming greedy longest palindrome picks all partitions
• Confusing counting palindromes with generating partitions
• Ignoring palindrome checks in BFS

2. You need to find the k-th lexicographically smallest permutation of numbers from 1 to n. Which approach guarantees an optimal solution without generating all permutations explicitly?

easy

A. Use a greedy algorithm that swaps elements to reach the k-th permutation directly.

B. Compute factorial values to determine each digit's position and build the permutation directly.

C. Use dynamic programming to count permutations and reconstruct the k-th sequence.

D. Generate all permutations using backtracking and return the k-th one.

Solution

1. Step 1: Understand the problem constraints

   The problem requires finding the k-th permutation without enumerating all permutations, which is infeasible for large n.

2. Step 2: Identify the approach that uses factorial number system

   Using precomputed factorials, we can determine the index of each digit in the permutation directly, avoiding full enumeration.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Factorial-based direct computation is optimal and avoids timeouts [OK]

Hint: Factorials let you jump directly to the k-th permutation [OK]

Common Mistakes:

• Thinking greedy swaps can find k-th permutation efficiently
• Assuming DP is needed here
• Trying to generate all permutations for large n

3. You need to generate all possible orderings of a list of unique integers. Which algorithmic approach guarantees that all permutations are generated without duplicates and with optimal time complexity?

easy

A. Backtracking with either a used array or in-place swapping to explore all permutations

B. Sorting the array repeatedly and collecting unique sequences

C. Dynamic programming to build permutations from smaller subproblems

D. Greedy algorithm that picks the next smallest unused element at each step

Solution

1. Step 1: Understand the problem requires all permutations

   Generating all permutations means enumerating every possible ordering of the input elements without duplicates.

2. Step 2: Identify the approach that systematically explores all orderings

   Backtracking with a used array or in-place swapping explores all permutations by recursively building sequences or swapping elements, ensuring no duplicates and covering all possibilities.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Backtracking is the standard method for generating all permutations [OK]

Hint: Backtracking systematically explores all permutations [OK]

Common Mistakes:

• Thinking greedy or sorting can generate all permutations without duplicates

4. Consider the following buggy code snippet for restoring IP addresses. Which line contains the subtle bug that causes invalid IP addresses to be included?

def restore_ip_addresses(s: str) -> List[str]:
    res = []
    stack = [(0, [])]
    while stack:
        start, path = stack.pop()
        if len(path) == 4:
            if start == len(s):
                res.append(".".join(path))
            continue
        remaining_segments = 4 - len(path)
        remaining_chars = len(s) - start
        if remaining_chars < remaining_segments or remaining_chars > 3 * remaining_segments:
            continue
        for length in range(1, 4):
            if start + length > len(s):
                break
            segment = s[start:start+length]
            # Bug: Missing check for leading zeros
            if length == 3 and int(segment) > 255:
                continue
            stack.append((start + length, path + [segment]))
    return res

medium

A. Line missing check for segments with leading zeros

B. Line checking if segment length is 3 and value > 255

C. Line pruning based on remaining characters and segments

D. Line appending new segment to stack

Solution

1. Step 1: Identify missing validation

   The code lacks a check to reject segments with leading zeros like '00' or '01', which are invalid.

2. Step 2: Confirm other checks are correct

   Check for segment > 255 and pruning are present and correct; appending to stack is correct.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Missing leading zero check allows invalid IP segments [OK]

Hint: Leading zero check is mandatory to avoid invalid IPs [OK]

Common Mistakes:

• Forgetting leading zero validation
• Misplacing pruning conditions

5. Suppose the problem is modified so that digits can be repeated any number of times (reuse allowed), and you want to generate all letter combinations of length n from the given digits. Which modification to the algorithm correctly handles this reuse scenario?

hard

A. Modify backtracking to allow choosing the same digit multiple times by not incrementing the index after each choice until length n is reached

B. Sort digits and generate combinations by picking letters only once per digit

C. Use dynamic programming to store combinations of length k and build up to length n without reuse

D. Use the same iterative queue approach but do not advance the digit index; instead, for each combination, append letters from all digits repeatedly until length n is reached

Solution

1. Step 1: Understand reuse requirement

   Allowing reuse means digits can be chosen multiple times to build combinations of length n.

2. Step 2: Identify correct approach

   Backtracking can be modified to not increment the digit index after choosing a letter, allowing reuse until length n is reached.

3. Step 3: Why other options fail

   Iterative queue approach as-is assumes fixed digit positions; DP without reuse does not handle repeated digits; sorting and picking once per digit ignores reuse.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Backtracking with flexible index handles reuse elegantly [OK]

Hint: Backtracking index controls reuse by not advancing [OK]

Common Mistakes:

• Trying to reuse digits in iterative approach without index control
• Ignoring reuse in DP or sorting