How do you solve Special Permutations on LeetCode?

Special Permutations (LeetCode #2741) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^n) time complexity and O(n * 2^n) space complexity. Key insight: Understanding the special condition is crucial for both brute-force and optimal approaches.

What is the brute force solution for Special Permutations?

The brute force approach for Special Permutations is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute-force approach generates all possible permutations of the given array and checks each one to see if it meets the special condition. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n * 2^n).

What algorithm or data structure is used to solve Special Permutations?

Special Permutations uses the following concepts: Array, Dynamic Programming, Bit Manipulation, Bitmask. The recommended patterns to study are: Dynamic Programming, Bit Manipulation.

What are the interview tips for Special Permutations?

Start with a brute-force solution to demonstrate understanding before optimizing. Explain your thought process clearly, especially when transitioning to a more complex solution. Practice writing clean and efficient code, as readability is important in interviews.

What are common mistakes when solving Special Permutations?

Not checking all possible pairs when validating permutations. Failing to handle large numbers correctly due to overflow.

#2741

Special Permutations

Medium

ArrayDynamic ProgrammingBit ManipulationBitmaskDynamic ProgrammingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n * 2^n)
Space	O(n)	O(n * 2^n)

💡

Intuition

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

Using dynamic programming with bit masking allows us to efficiently track which numbers have been used in the permutation and the last number added, significantly reducing the number of checks needed.

⚙️

Algorithm

3 steps

1Step 1: Create a DP table where dp[mask][last] represents the number of special permutations using the numbers represented by 'mask' with 'last' as the last number in the permutation.
2Step 2: Iterate through all possible masks and for each mask, try to add each unused number to the permutation, checking the special condition.
3Step 3: Sum all valid permutations from the DP table and return the result modulo 10^9 + 7.

solution.py20 lines

1# Full working Python code
2def special_permutations(nums):
3    n = len(nums)
4    dp = [[0] * n for _ in range(1 << n)]
5    MOD = 10**9 + 7
6
7    for i in range(n):
8        dp[1 << i][i] = 1
9
10    for mask in range(1 << n):
11        for last in range(n):
12            if dp[mask][last] == 0:
13                continue
14            for next in range(n):
15                if mask & (1 << next):
16                    continue
17                if nums[last] % nums[next] == 0 or nums[next] % nums[last] == 0:
18                    dp[mask | (1 << next)][next] = (dp[mask | (1 << next)][next] + dp[mask][last]) % MOD
19
20    return sum(dp[(1 << n) - 1]) % MOD

ℹ

Complexity note: The time complexity is O(n * 2^n) because we explore all subsets of the numbers (2^n) and for each subset, we check each number (n). The space complexity is O(n * 2^n) for storing the DP table.

1Understanding the special condition is crucial for both brute-force and optimal approaches.
2Using bit masking allows us to efficiently track used numbers and avoid redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.