How do you solve Concatenated Divisibility on LeetCode?

Concatenated Divisibility (LeetCode #3533) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^n) time complexity and O(2^n) space complexity. Key insight: Permutations can be costly; use bitmasks to optimize.

What is the time complexity of Concatenated Divisibility?

The optimal solution for Concatenated Divisibility runs in O(n * 2^n) time and uses O(2^n) space. Each state in DP is computed in O(n) time, and there are 2^n states.

What is the brute force solution for Concatenated Divisibility?

The brute force approach for Concatenated Divisibility is Brute Force, with O(n!) time complexity and O(n) space complexity. Generate all permutations of the array and check each concatenated number for divisibility by k. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n * 2^n).

What algorithm or data structure is used to solve Concatenated Divisibility?

Concatenated Divisibility uses the following concepts: Array, Dynamic Programming, Bit Manipulation, Bitmask. The recommended patterns to study are: Backtracking, Dynamic Programming.

What are the interview tips for Concatenated Divisibility?

Explain your thought process clearly. Discuss edge cases like single-element arrays. Practice generating permutations and bitmasking.

What are common mistakes when solving Concatenated Divisibility?

Not considering leading zeros in concatenation. Forgetting to sort the final result.

#3533

Concatenated Divisibility

Hard

ArrayDynamic ProgrammingBit ManipulationBitmaskBacktrackingDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

Use dynamic programming with bitmasks to track used numbers and their concatenated value's modulus. This reduces the number of checks significantly.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP table where dp[mask] holds the smallest concatenated number mod k for the bitmask 'mask'.
2Step 2: Iterate through all masks and for each number not in the mask, update the new mask and calculate the new mod value.
3Step 3: After filling the DP table, find the smallest permutation that gives a mod value of 0.

solution.py17 lines

1from itertools import permutations
2
3def divisible_permutation(nums, k):
4    n = len(nums)
5    dp = [None] * (1 << n)
6    dp[0] = ''
7    for mask in range(1 << n):
8        for i in range(n):
9            if not (mask & (1 << i)):
10                new_mask = mask | (1 << i)
11                new_num = dp[mask] + str(nums[i]) if dp[mask] is not None else str(nums[i])
12                if dp[new_mask] is None or new_num < dp[new_mask]:
13                    dp[new_mask] = new_num
14    for mask in range(1 << n):
15        if dp[mask] is not None and int(dp[mask]) % k == 0:
16            return sorted([nums[i] for i in range(n) if mask & (1 << i)])
17    return []

ℹ

Complexity note: Each state in DP is computed in O(n) time, and there are 2^n states.

1Permutations can be costly; use bitmasks to optimize.
2Divisibility can be checked using modular arithmetic.

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