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3352. Count K-Reducible Numbers Less Than N 👍

  • Time: $O(n^2)$
  • Space: $O(n)$
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class Solution {
 public:
  int countKReducibleNumbers(string s, int k) {
    vector<vector<vector<int>>> mem(
        s.length(), vector<vector<int>>(s.length() + 1, vector<int>(2, -1)));
    return count(s, 0, 0, true, k, getOps(s), mem) - 1;  // - 0
  }

 private:
  static constexpr int kMod = 1'000'000'007;

  // Returns the number of positive integers less than n that are k-reducible,
  // considering the i-th digit, where `setBits` is the number of set bits in
  // the current number, and `tight` indicates if the current digit is
  // tightly bound.
  int count(const string& s, int i, int setBits, bool tight, int k,
            const vector<int>& ops, vector<vector<vector<int>>>& mem) {
    if (i == s.length())
      return (ops[setBits] < k && !tight) ? 1 : 0;
    if (mem[i][setBits][tight] != -1)
      return mem[i][setBits][tight];

    int res = 0;
    const int maxDigit = tight ? s[i] - '0' : 1;

    for (int d = 0; d <= maxDigit; ++d) {
      const bool nextTight = tight && (d == maxDigit);
      res += count(s, i + 1, setBits + d, nextTight, k, ops, mem);
      res %= kMod;
    }

    return mem[i][setBits][tight] = res;
  };

  // Returns the number of operations to reduce a number to 0.
  vector<int> getOps(string& s) {
    vector<int> ops(s.length() + 1);
    for (int num = 2; num <= s.length(); ++num)
      ops[num] = 1 + ops[__builtin_popcount(num)];
    return ops;
  }
};

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class Solution {
  public int countKReducibleNumbers(String s, int k) {
    Integer[][][] mem = new Integer[s.length()][s.length() + 1][2];
    return count(s, 0, 0, true, k, getOps(s), mem) - 1; // - 0
  }

  private static final int MOD = 1_000_000_007;

  // Returns the number of positive integers less than n that are k-reducible,
  // considering the i-th digit, where `setBits` is the number of set bits in
  // the current number, and `tight` indicates if the current digit is
  // tightly bound.
  private int count(String s, int i, int setBits, boolean tight, int k, int[] ops,
                    Integer[][][] mem) {
    if (i == s.length())
      return (ops[setBits] < k && !tight) ? 1 : 0;
    if (mem[i][setBits][tight ? 1 : 0] != null)
      return mem[i][setBits][tight ? 1 : 0];

    int res = 0;
    final int maxDigit = tight ? s.charAt(i) - '0' : 1;

    for (int d = 0; d <= maxDigit; ++d) {
      final boolean nextTight = tight && (d == maxDigit);
      res += count(s, i + 1, setBits + d, nextTight, k, ops, mem);
      res %= MOD;
    }

    return mem[i][setBits][tight ? 1 : 0] = res;
  }

  // Returns the number of operations to reduce a number to 0.
  private int[] getOps(String s) {
    int[] ops = new int[s.length() + 1];
    for (int num = 2; num <= s.length(); ++num)
      ops[num] = 1 + ops[Integer.bitCount(num)];
    return ops;
  }
}

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class Solution:
  def countKReducibleNumbers(self, s: str, k: int) -> int:
    MOD = 1_000_000_007
    ops = self._getOps(s)

    @functools.lru_cache(None)
    def dp(i: int, setBits: int, tight: bool) -> int:
      """
      Returns the number of positive integers less than n that are k-reducible,
      considering the i-th digit, where `setBits` is the number of set bits in
      the current number, and `tight` indicates if the current digit is
      tightly bound.
      """
      if i == len(s):
        return int(ops[setBits] < k and not tight)

      res = 0
      maxDigit = int(s[i]) if tight else 1

      for d in range(maxDigit + 1):
        nextTight = tight and (d == maxDigit)
        res += dp(i + 1, setBits + d, nextTight)
        res %= MOD
      return res

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

  def _getOps(self, s: str) -> int:
    """Returns the number of operations to reduce a number to 0."""
    ops = [0] * (len(s) + 1)
    for num in range(2, len(s) + 1):
      ops[num] = 1 + ops[num.bit_count()]
    return ops