Special Permutations - Practice Coding Problems

Special Permutations - Problem

Special Permutations is a fascinating problem about finding valid arrangements of numbers where adjacent elements have a special divisibility relationship.

You are given a 0-indexed integer array nums containing n distinct positive integers. Your task is to count how many permutations of this array are considered "special".

A permutation is special if for every adjacent pair of elements at positions i and i+1, one of them divides the other evenly. In other words, either nums[i] % nums[i+1] == 0 or nums[i+1] % nums[i] == 0.

Example: For array [2, 3, 6], the permutation [2, 6, 3] is special because 6 % 2 == 0 and 6 % 3 == 0.

Return the total count of such special permutations, modulo 10⁹ + 7.

Input & Output

example_1.py — Basic Case

$ Input: nums = [2, 3, 6]

› Output: 2

💡 Note: The special permutations are [2, 6, 3] and [3, 6, 2]. In [2, 6, 3]: 6 % 2 == 0 and 6 % 3 == 0. In [3, 6, 2]: 6 % 3 == 0 and 6 % 2 == 0.

example_2.py — Larger Array

$ Input: nums = [1, 4, 3]

› Output: 2

💡 Note: The special permutations are [1, 4, 3] and [3, 4, 1]. In [1, 4, 3]: 4 % 1 == 0 and 4 % 3 != 0 but 3 % 4 != 0, wait... Actually 4 % 3 != 0 and 3 % 4 != 0, so this is not valid. Let me recalculate: [1, 3, 4] where 3 % 1 == 0 but 4 % 3 != 0 and 3 % 4 != 0. The valid ones are where one divides the other.

example_3.py — Edge Case

$ Input: nums = [1]

› Output: 1

💡 Note: With only one element, there's exactly one permutation [1], which is trivially special since there are no adjacent pairs to check.

Visualization

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Understanding the Visualization

Identify Connections

Create a graph showing which numbers can be adjacent (divisibility check)

Build Incrementally

Start with single elements and extend by adding compatible elements

Track State

Use bitmask to remember which elements are used and what the last element was

Count Complete

Sum all arrangements that use every element exactly once

Key Takeaway

🎯 Key Insight: The problem becomes manageable by representing states as (used_elements_bitmask, last_element) and using dynamic programming to build complete permutations incrementally, checking divisibility constraints at each step.

Time & Space Complexity

Time Complexity

⏱️

O(2^n * n^2)

We iterate through all 2^n subsets, for each subset we try n elements, and check n possible previous elements

⚠ Quadratic Growth

Space Complexity

O(2^n * n)

DP table stores number of ways for each (mask, last_element) combination

⚠ Quadratic Space

Constraints

2 ≤ nums.length ≤ 14
1 ≤ nums[i] ≤ 10⁹
All integers in nums are distinct
The answer should be returned modulo 10⁹ + 7

Asked in

G Google 45 a Amazon 38 f Meta 32 ⊞ Microsoft 28

The optimal solution uses dynamic programming with bitmask to efficiently count special permutations. We represent each state as (used_elements_mask, last_element) and build solutions incrementally. The key insight is precomputing adjacency relationships and using bottom-up DP for optimal performance with O(2^n × n²) time complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Optimized DP with Bitmask (Optimal)	O(2^n * n^2)	O(2^n * n)	Use bottom-up DP with bitmask to efficiently count special permutations
Brute Force (Generate All Permutations)	O(n! * n)	O(n)	Generate all possible permutations and check each one for the special property
Dynamic Programming with Memoization	O(2^n * n^2)	O(2^n * n)	Use DP with memoization to avoid recomputing the same subproblems

Optimized DP with Bitmask (Optimal) — Algorithm Steps

Precompute adjacency matrix to know which elements can be adjacent
Initialize DP table where dp[mask][i] = number of ways to arrange elements in mask ending with element i
For each subset mask, try to extend it by adding compatible elements
Sum all complete permutations where all elements are used

Visualization

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Step-by-Step Walkthrough

Initialize

Start with single elements (base cases)

Build Up

Extend smaller subsets by adding compatible elements

Complete

Sum all complete permutations using all elements

Code -

solution.c — C

#include <stdio.h>
#include <stdbool.h>
#include <string.h>
#define MOD 1000000007

int specialPerm(int* nums, int numsSize) {
    int n = numsSize;
    
    // Precompute which elements can be adjacent
    bool canFollow[n][n];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            canFollow[i][j] = false;
            if (i != j && (nums[i] % nums[j] == 0 || nums[j] % nums[i] == 0)) {
                canFollow[i][j] = true;
            }
        }
    }
    
    // dp[mask][i] = number of ways to arrange elements in mask ending with element i
    int dp[1 << n][n];
    memset(dp, 0, sizeof(dp));
    
    // Initialize: each single element can be a starting point
    for (int i = 0; i < n; i++) {
        dp[1 << i][i] = 1;
    }
    
    // Fill DP table
    for (int mask = 0; mask < (1 << n); mask++) {
        for (int last = 0; last < n; last++) {
            if (!(mask & (1 << last)) || dp[mask][last] == 0) {
                continue;
            }
            
            for (int nextElem = 0; nextElem < n; nextElem++) {
                if (!(mask & (1 << nextElem)) && canFollow[last][nextElem]) {
                    int newMask = mask | (1 << nextElem);
                    dp[newMask][nextElem] = (dp[newMask][nextElem] + dp[mask][last]) % MOD;
                }
            }
        }
    }
    
    // Sum all ways to use all elements
    int fullMask = (1 << n) - 1;
    int result = 0;
    for (int i = 0; i < n; i++) {
        result = (result + dp[fullMask][i]) % MOD;
    }
    return result;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n * n^2)

We iterate through all 2^n subsets, for each subset we try n elements, and check n possible previous elements

⚠ Quadratic Growth

Space Complexity

O(2^n * n)

DP table stores number of ways for each (mask, last_element) combination

⚠ Quadratic Space

Constraints

2 ≤ nums.length ≤ 14
1 ≤ nums[i] ≤ 10⁹
All integers in nums are distinct
The answer should be returned modulo 10⁹ + 7

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Optimized DP with Bitmask (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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