How do you solve Check Digitorial Permutation on LeetCode?

Check Digitorial Permutation (LeetCode #3848) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Factorials grow quickly, limiting possible digitorial numbers.

What is the time complexity of Check Digitorial Permutation?

The optimal solution for Check Digitorial Permutation runs in O(n) time and uses O(1) space. The complexity is linear as we only compute factorials once and sum them in a single pass.

What is the brute force solution for Check Digitorial Permutation?

The brute force approach for Check Digitorial Permutation is Brute Force, with O(n!) time complexity and O(n) space complexity. Check all permutations of the digits of n and see if any permutation is a digitorial number. This is inefficient due to the factorial growth of permutations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check Digitorial Permutation?

Check Digitorial Permutation uses the following concepts: Math, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Check Digitorial Permutation?

Explain your thought process clearly. Discuss edge cases, like single-digit numbers. Practice with similar problems to build confidence.

What are common mistakes when solving Check Digitorial Permutation?

Forgetting to check leading zeros in permutations. Not precomputing factorials, leading to repeated calculations.

#3848

Check Digitorial Permutation

Medium

MathCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n)
Space	O(n)	O(1)

💡

Intuition

Time O(n)Space O(1)

Precompute the factorials of digits 0-9, then calculate the sum of factorials of the digits of n. Check if this sum can be formed using the digits of n.

⚙️

Algorithm

3 steps

1Step 1: Precompute factorials for digits 0-9.
2Step 2: Calculate the sum of the factorials of the digits of n.
3Step 3: Check if the digits of the sum match the digits of n (no leading zero).

solution.py6 lines

1from collections import Counter
2
3def is_digitorial(n):
4    factorials = [1] + [i * factorials[i-1] for i in range(1, 10)]
5    digit_sum = sum(factorials[int(d)] for d in str(n))
6    return Counter(str(digit_sum)) == Counter(str(n)) and str(digit_sum)[0] != '0'

ℹ

Complexity note: The complexity is linear as we only compute factorials once and sum them in a single pass.

1Factorials grow quickly, limiting possible digitorial numbers.
2Only digits 0-9 matter for factorial calculations.

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