Solution Approach

The implementation follows these steps:

1. Precompute Factorials

fac = [factorial(i) for i in range(n + 1)]

We precompute factorials from 0! to n! for efficient permutation counting later.

2. Initialize Variables

• ans: Counter for the final result
• vis: Set to track processed digit combinations (avoids duplicates)
• base = 10^((n-1)//2): Starting point for enumerating the first half of palindromes

3. Generate Palindromes

for i in range(base, base * 10):  s = str(i)  s += s[::-1][n % 2:]

For each number i in the range [base, base * 10):

• Convert it to string s
• Append its reverse to form a palindrome
• If n is odd (n % 2 = 1), skip the first character of the reverse to avoid duplicating the middle digit
• If n is even (n % 2 = 0), append the full reverse

Example: If n = 5 and i = 123, we get s = "123" + "21" = "12321"

4. Check Divisibility by k

if int(s) % k:  continue

Skip palindromes that aren't divisible by k.

5. Handle Duplicate Digit Sets

t = "".join(sorted(s)) if t in vis:  continue vis.add(t)

Sort the digits of the palindrome to create a unique identifier. If we've already processed this digit combination, skip it. Otherwise, add it to vis.

6. Count Valid Permutations

cnt = Counter(t) res = (n - cnt["0"]) * fac[n - 1] for x in cnt.values():  res //= fac[x] ans += res

Using combinatorics formula: $\frac{(n - x_0) \cdot (n - 1)!}{x_0! \cdot x_1! \cdots x_9!}$

• Counter(t) counts the frequency of each digit
• (n - cnt["0"]) represents valid choices for the first position (non-zero digits)
• fac[n - 1] represents (n-1)! permutations for remaining positions
• Divide by fac[x] for each digit frequency to eliminate duplicate permutations
• Add the result to the total answer

The algorithm efficiently counts all n-digit good integers by generating palindromes, checking divisibility, and using combinatorics to count valid permutations while avoiding duplicates.

Example Walkthrough

Let's walk through the solution with n = 4 and k = 2.

Step 1: Setup

• Precompute factorials: fac = [1, 1, 2, 6, 24] (for 0! through 4!)
• Initialize ans = 0, vis = set()
• Calculate base = 10^((4-1)//2) = 10^1 = 10

Step 2: Generate 4-digit palindromes We iterate i from 10 to 99 (first half of 4-digit palindromes):

For i = 10:

• s = "10" + "01" = "1001" (palindrome)
• Check: 1001 % 2 = 1 (odd, not divisible by 2)
• Skip this palindrome

For i = 11:

• s = "11" + "11" = "1111" (palindrome)
• Check: 1111 % 2 = 1 (odd, not divisible by 2)
• Skip this palindrome

For i = 12:

• s = "12" + "21" = "1221" (palindrome)
• Check: 1221 % 2 = 1 (odd, not divisible by 2)
• Skip this palindrome

...continuing...

For i = 20:

• s = "20" + "02" = "2002" (palindrome)
• Check: 2002 % 2 = 0 ✓ (divisible by 2)
• Sort digits: t = "0022"
• Not in vis, so add "0022" to vis
• Count permutations:
  • cnt = Counter("0022") = {'0': 2, '2': 2}
  • Valid first positions: 4 - 2 = 2 (can't use zeros)
  • Permutations: res = 2 * 3! / (2! * 2!) = 2 * 6 / 4 = 3
  • The 3 valid numbers are: 2002, 2020, 2200
• Add to answer: ans = 0 + 3 = 3

For i = 22:

• s = "22" + "22" = "2222" (palindrome)
• Check: 2222 % 2 = 0 ✓ (divisible by 2)
• Sort digits: t = "2222"
• Not in vis, so add "2222" to vis
• Count permutations:
  • cnt = Counter("2222") = {'2': 4}
  • Valid first positions: 4 - 0 = 4 (no zeros)
  • Permutations: res = 4 * 3! / 4! = 4 * 6 / 24 = 1
  • The only valid number is: 2222
• Add to answer: ans = 3 + 1 = 4

...continuing through all values...

For i = 40:

• s = "40" + "04" = "4004" (palindrome)
• Check: 4004 % 2 = 0 ✓ (divisible by 2)
• Sort digits: t = "0044"
• Not in vis, so add "0044" to vis
• Count permutations:
  • cnt = Counter("0044") = {'0': 2, '4': 2}
  • Valid first positions: 4 - 2 = 2
  • Permutations: res = 2 * 3! / (2! * 2!) = 2 * 6 / 4 = 3
  • The 3 valid numbers are: 4004, 4040, 4400
• Add to answer: ans = 4 + 3 = 7

The process continues for all even palindromes from 10 to 99, accumulating the count of good integers. Each valid k-palindromic number contributes the count of its unique permutations (excluding those with leading zeros) to the final answer.

Time and Space Complexity

Time Complexity: O(10^m × n × log n) where m = ⌊(n-1)/2⌋

• The main loop iterates from base to base * 10, where base = 10^((n-1)//2). This gives us 10^m iterations where m = ⌊(n-1)/2⌋.
• For each iteration:
  • String concatenation and reversal operations: O(n) (since the resulting string has length n)
  • Modulo operation on an n-digit number: O(n)
  • Sorting the string s to create t: O(n log n)
  • Checking membership in vis set: O(n) (comparing strings of length n)
  • Creating Counter from string t: O(n)
  • Computing factorial division for permutations: O(n) (at most n distinct digits)
• The dominant operation per iteration is sorting: O(n log n)
• Total: O(10^m × n × log n)

Space Complexity: O(10^m × n) where m = ⌊(n-1)/2⌋

• fac array stores n+1 factorial values: O(n)
• vis set can store at most 10^m unique sorted strings, each of length n: O(10^m × n)
• Other variables like s, t, and cnt use O(n) space but are reused in each iteration
• The dominant space usage is the vis set: O(10^m × n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Palindrome Generation for Odd-Length Numbers

Pitfall: When generating palindromes of odd length, a common mistake is duplicating the middle digit incorrectly. For example, if n = 5 and we start with 123, we might incorrectly generate 12332321 (8 digits) instead of 12321 (5 digits).

Why it happens: The slicing logic s[::-1][n % 2:] can be confusing. When n is odd, we need to skip the first character of the reversed string to avoid duplicating the middle digit.

Solution: Ensure the palindrome generation logic correctly handles both odd and even lengths:

# For odd n: append reverse without the first character # For even n: append the full reverse palindrome_str = half_str + half_str[::-1][n % 2:]

2. Division by Zero in Permutation Calculation

Pitfall: If the factorial array isn't properly initialized or if digit counts are incorrect, you might attempt to divide by factorial[0] when it's undefined or zero.

Why it happens: The code divides by factorials[count] for each digit count. If factorials[0] isn't properly set to 1, this will cause an error.

Solution: Always initialize factorial(0) = 1:

factorials = [factorial(i) for i in range(n + 1)] # factorial(0) = 1 by definition

3. Integer Overflow in Permutation Counting

Pitfall: For large values of n, the intermediate calculation (n - digit_counts["0"]) * factorials[n - 1] can overflow, even though the final result after division might be within bounds.

Why it happens: Python handles big integers well, but in other languages or if optimizing, you might face overflow issues.

Solution: Perform divisions as early as possible to keep numbers manageable:

# Instead of computing the full numerator first valid_permutations = factorials[n - 1] for count in digit_counts.values():  valid_permutations //= factorials[count] valid_permutations *= (n - digit_counts["0"])

4. Missing Edge Case: All Zeros in Digit Count

Pitfall: When a palindrome consists entirely of zeros (which shouldn't happen for valid n-digit numbers), the formula (n - digit_counts["0"]) would be 0, but this case should never occur since we're generating valid palindromes.

Why it happens: The palindrome generation starts from 10^((n-1)//2), which ensures at least one non-zero digit, but it's worth being aware of this constraint.

Solution: The current implementation naturally avoids this by starting palindrome generation from a non-zero base:

half_start = 10 ** ((n - 1) // 2) # Always starts with a non-zero digit

5. Incorrect Handling of Single-Digit Numbers (n=1)

Pitfall: When n = 1, the calculation 10 ** ((n - 1) // 2) equals 10 ** 0 = 1, which means the range would be [1, 10). This works correctly, but the permutation counting logic might seem unnecessary for single digits.

Solution: The current implementation handles this correctly, but you could add a special case for clarity:

if n == 1:  # Count single-digit numbers divisible by k  return len([i for i in range(1, 10) if i % k == 0])