LeetCode Solutions

			

class Solution {
 public:
  int numDupDigitsAtMostN(int n) {
    return n - countSpecialNumbers(n);
  }

 private:
  // Same as 2376. Count Special Integers
  int countSpecialNumbers(int n) {
    const int digitSize = log10(n) + 1;
    // dp[i][j][k] := # of special integers that belong to the interval
    // [0, 10^i] with `usedMask` j, where k is 0/1 tight constraint
    dp.resize(digitSize + 1, vector<vector<int>>(1 << 10, vector<int>(2, -1)));
    return count(to_string(n), digitSize, 0, true) - 1;  // - 0;
  }

  vector<vector<vector<int>>> dp;

  int count(const string& s, int digitSize, int usedMask, bool isTight) {
    if (digitSize == 0)
      return 1;
    if (dp[digitSize][usedMask][isTight] != -1)
      return dp[digitSize][usedMask][isTight];

    int ans = 0;
    const int maxDigit = isTight ? s[s.length() - digitSize] - '0' : 9;

    for (int digit = 0; digit <= maxDigit; ++digit) {
      // `digit` is used
      if (usedMask >> digit & 1)
        continue;
      // Use `digit` now
      const bool nextIsTight = isTight && (digit == maxDigit);
      if (usedMask == 0 && digit == 0)  // don't count leading 0s as used
        ans += count(s, digitSize - 1, usedMask, nextIsTight);
      else
        ans += count(s, digitSize - 1, usedMask | 1 << digit, nextIsTight);
    }

    return dp[digitSize][usedMask][isTight] = ans;
  }
};

			

class Solution {
  public int numDupDigitsAtMostN(int n) {
    return n - countSpecialNumbers(n);
  }

  // Same as 2376. Count Special Integers
  private int countSpecialNumbers(int n) {
    final int digitSize = (int) Math.log10(n) + 1;
    // dp[i][j][k] := # of special integers that belong to the interval
    // [0, 10^i] with `usedMask` j, where k is 0/1 tight constraint
    dp = new Integer[digitSize + 1][1 << 10][2];
    return count(String.valueOf(n), digitSize, 0, true) - 1; // - 0;
  }

  private Integer[][][] dp;

  private int count(final String s, int digitSize, int usedMask, boolean isTight) {
    if (digitSize == 0)
      return 1;
    if (dp[digitSize][usedMask][isTight ? 1 : 0] != null)
      return dp[digitSize][usedMask][isTight ? 1 : 0];

    int ans = 0;
    final int maxDigit = isTight ? s.charAt(s.length() - digitSize) - '0' : 9;

    for (int digit = 0; digit <= maxDigit; ++digit) {
      // `digit` is used
      if ((usedMask >> digit & 1) == 1)
        continue;
      // Use `digit` now
      final boolean nextIsTight = isTight && (digit == maxDigit);
      if (usedMask == 0 && digit == 0) // don't count leading 0s as used
        ans += count(s, digitSize - 1, usedMask, nextIsTight);
      else
        ans += count(s, digitSize - 1, usedMask | 1 << digit, nextIsTight);
    }

    return dp[digitSize][usedMask][isTight ? 1 : 0] = ans;
  }
}

			

class Solution:
  def numDupDigitsAtMostN(self, n: int) -> int:
    return n - self._countSpecialNumbers(n)

  def _countSpecialNumbers(self, n: int) -> int:
    s = str(n)
    digitSize = int(log10(n)) + 1

    # Dp(i, j, k) := # Of special integers that beto the interval
    # [0, 10^i] with `usedMask` j, where k is 0/1 tight constraint
    @functools.lru_cache(None)
    def dp(digitSize: int, usedMask: int, isTight: bool) -> int:
      if digitSize == 0:
        return 1

      ans = 0
      maxDigit = ord(s[len(s) - digitSize]) - ord('0') if isTight else 9

      for digit in range(maxDigit + 1):
        # `digit` is used
        if usedMask >> digit & 1:
          continue
        # Use `digit` now
        nextIsTight = isTight and (digit == maxDigit)
        if usedMask == 0 and digit == 0:  # don't count leading 0s as used
          ans += dp(digitSize - 1, usedMask, nextIsTight)
        else:
          ans += dp(digitSize - 1, usedMask | 1 << digit, nextIsTight)

      return ans

    return dp(digitSize, 0, True) - 1  # - 0