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3272. Find the Count of Good Integers 👍

  • Time: $O(n \cdot 10^{n / 2})$
  • Space: $O(n)$
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class Solution {
 public:
  long long countGoodIntegers(int n, int k) {
    const int halfLength = (n + 1) / 2;
    const int minHalf = pow(10, halfLength - 1);
    const int maxHalf = pow(10, halfLength);
    long ans = 0;
    unordered_set<string> seen;

    for (int num = minHalf; num < maxHalf; ++num) {
      const string firstHalf = to_string(num);
      const string secondHalf = {firstHalf.rbegin(), firstHalf.rend()};
      const string palindrome = firstHalf + secondHalf.substr(n % 2);
      if (stol(palindrome) % k != 0)
        continue;
      string sortedDigits = palindrome;
      ranges::sort(sortedDigits);
      if (seen.contains(sortedDigits))
        continue;
      seen.insert(sortedDigits);
      vector<int> digitCount(10);
      for (const char c : palindrome)
        ++digitCount[c - '0'];
      // Leading zeros are not allowed, so the first digit is special.
      const int firstDigitChoices = n - digitCount[0];
      long permutations = firstDigitChoices * factorial(n - 1);
      // For each repeated digit, divide by the factorial of the frequency since
      // permutations that swap identical digits don't create a new number.
      for (const int freq : digitCount)
        if (freq > 1)
          permutations /= factorial(freq);
      ans += permutations;
    }

    return ans;
  }

 private:
  long factorial(int n) {
    long res = 1;
    for (int i = 2; i <= n; ++i)
      res *= i;
    return res;
  }
};

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class Solution {
  public long countGoodIntegers(int n, int k) {
    final int halfLength = (n + 1) / 2;
    final int minHalf = (int) Math.pow(10, halfLength - 1);
    final int maxHalf = (int) Math.pow(10, halfLength);
    long ans = 0;
    Set<String> seen = new HashSet<>();

    for (int num = minHalf; num < maxHalf; ++num) {
      final String firstHalf = String.valueOf(num);
      final String secondHalf = new StringBuilder(firstHalf).reverse().toString();
      final String palindrome = firstHalf + secondHalf.substring(n % 2);
      if (Long.parseLong(palindrome) % k != 0)
        continue;
      char[] sortedDigits = palindrome.toCharArray();
      Arrays.sort(sortedDigits);
      String sortedDigitsStr = new String(sortedDigits);
      if (seen.contains(sortedDigitsStr))
        continue;
      seen.add(sortedDigitsStr);
      int[] digitCount = new int[10];
      for (char c : palindrome.toCharArray())
        ++digitCount[c - '0'];
      // Leading zeros are not allowed, so the first digit is special.
      final int firstDigitChoices = n - digitCount[0];
      long permutations = firstDigitChoices * factorial(n - 1);
      // For each repeated digit, divide by the factorial of the frequency since
      // permutations that swap identical digits don't create a new number.
      for (final int freq : digitCount)
        if (freq > 1)
          permutations /= factorial(freq);
      ans += permutations;
    }

    return ans;
  }

  private long factorial(int n) {
    long res = 1;
    for (int i = 2; i <= n; ++i)
      res *= i;
    return res;
  }
}

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class Solution:
  def countGoodIntegers(self, n: int, k: int) -> int:
    halfLength = (n + 1) // 2
    minHalf = 10**(halfLength - 1)
    maxHalf = 10**halfLength
    ans = 0
    seen = set()

    for num in range(minHalf, maxHalf):
      palindrome = str(num) + str(num)[::-1][n % 2:]
      sortedDigits = ''.join(sorted(palindrome))
      if int(palindrome) % k != 0 or sortedDigits in seen:
        continue
      seen.add(sortedDigits)
      digitCount = collections.Counter(palindrome)
      # Leading zeros are not allowed, so the first digit is special.
      firstDigitChoices = n - digitCount['0']
      permutations = firstDigitChoices * math.factorial(n - 1)
      # For each repeated digit, divide by the factorial of the frequency since
      # permutations that swap identical digits don't create a new number.
      for freq in digitCount.values():
        permutations //= math.factorial(freq)
      ans += permutations

    return ans