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2539. Count the Number of Good Subsequences 👎

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  int countGoodSubsequences(string s) {
    // For each frequency f in [1, max(freq)], start with "" and calculate how
    // many subsequences can be constructed with each letter's frequency = f.
    //
    // e.g. s = "abb", so f = max(freq) = 2.
    //
    // For f = 1, with 1 way to build "", choose any 'a' to construct a good
    // subseq, so # of good subsequences = 1 + 1 * (1, 1) = 2 ("", "a"). Next,
    // add 'b' and # of good subsequences = 2 + 2 * (2, 1) = 6 ("", "a", "b1",
    // "b2", "ab1", "ab2"). So, the number of good subsequences for f = 1 is 5
    // since we need to exclude "".
    //
    // For f = 2, with 1 way to build "", choose any two 'b's to construct a
    // good subseq, so # of good subsequences = 1 + 1 * (2, 2) is 2 ("", "bb").
    // So, the number of good subsequences for f = 2 = 1 since we need to
    // exclude "".
    //
    // Therefore, the number of good subsequences for "aab" = 5 + 1 = 6.
    int ans = 0;
    vector<int> count(26);

    for (const char c : s)
      ++count[c - 'a'];

    const int maxFreq = ranges::max(count);
    const auto [fact, invFact] = getFactAndInvFact(maxFreq);

    for (int freq = 1; freq <= maxFreq; ++freq) {
      long numSubseqs = 1;  // ""
      for (const int charFreq : count)
        if (charFreq >= freq)
          numSubseqs = (numSubseqs +  //
                        numSubseqs * nCk(charFreq, freq, fact, invFact)) %
                       kMod;
      ans += numSubseqs - 1;  // Minus "".
      ans %= kMod;
    }

    return ans;
  }

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

  pair<vector<long>, vector<long>> getFactAndInvFact(int n) {
    vector<long> fact(n + 1);
    vector<long> invFact(n + 1);
    vector<long> inv(n + 1);
    fact[0] = invFact[0] = 1;
    inv[0] = inv[1] = 1;
    for (int i = 1; i <= n; ++i) {
      if (i >= 2)
        inv[i] = kMod - kMod / i * inv[kMod % i] % kMod;
      fact[i] = fact[i - 1] * i % kMod;
      invFact[i] = invFact[i - 1] * inv[i] % kMod;
    }
    return {fact, invFact};
  }

  int nCk(int n, int k, const vector<long>& fact, const vector<long>& invFact) {
    return fact[n] * invFact[k] % kMod * invFact[n - k] % kMod;
  }
};

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class Solution {
  public int countGoodSubsequences(String s) {
    // For each frequency f in [1, max(freq)], start with "" and calculate how
    // many subsequences can be finalructed with each letter's frequency = f.
    //
    // e.g. s = "abb", so f = max(freq) = 2.
    //
    // For f = 1, with 1 way to build "", choose any 'a' to finalruct a good
    // subseq, so # of good subsequences = 1 + 1 * (1, 1) = 2 ("", "a"). Next, add
    // 'b' and # of good subsequences = 2 + 2 * (2, 1) = 6 ("", "a", "b1", "b2",
    // "ab1", "ab2"). So, the number of good subsequences for f = 1 is 5 since we need to
    // exclude "".
    //
    // For f = 2, with 1 way to build "", choose any two 'b's to finalruct a
    // good subseq, so # of good subsequences = 1 + 1 * (2, 2) is 2 ("", "bb"). So,
    // the number of good subsequences for f = 2 = 1 since we need to exclude "".
    //
    // Therefore, the number of good subsequences for "aab" = 5 + 1 = 6.
    final int kMod = 1_000_000_007;
    int ans = 0;
    int[] count = new int[26];

    for (final char c : s.toCharArray())
      ++count[c - 'a'];

    final int maxFreq = Arrays.stream(count).max().getAsInt();
    final long[][] factAndInvFact = getFactAndInvFact(maxFreq);
    final long[] fact = factAndInvFact[0];
    final long[] invFact = factAndInvFact[1];

    for (int freq = 1; freq <= maxFreq; ++freq) {
      long numSubseqs = 1; // ""
      for (final int charFreq : count)
        if (charFreq >= freq)
          numSubseqs = (numSubseqs + //
                        numSubseqs * nCk(charFreq, freq, fact, invFact)) %
                       kMod;
      ans += numSubseqs - 1; // Minus "".
      ans %= kMod;
    }

    return ans;
  }

  private static final int kMod = 1_000_000_007;

  private long[][] getFactAndInvFact(int n) {
    long[] fact = new long[n + 1];
    long[] invFact = new long[n + 1];
    long[] inv = new long[n + 1];
    fact[0] = invFact[0] = 1;
    inv[0] = inv[1] = 1;
    for (int i = 1; i <= n; ++i) {
      if (i >= 2)
        inv[i] = kMod - kMod / i * inv[kMod % i] % kMod;
      fact[i] = fact[i - 1] * i % kMod;
      invFact[i] = invFact[i - 1] * inv[i] % kMod;
    }
    return new long[][] {fact, invFact};
  }

  private int nCk(int n, int k, long[] fact, long[] invFact) {
    return (int) (fact[n] * invFact[k] % kMod * invFact[n - k] % kMod);
  }
}

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class Solution:
  def countGoodSubsequences(self, s: str) -> int:
    kMod = 1_000_000_007
    ans = 0
    count = collections.Counter(s)

    @functools.lru_cache(None)
    def fact(i: int) -> int:
      return 1 if i <= 1 else i * fact(i - 1) % kMod

    @functools.lru_cache(None)
    def inv(i: int) -> int:
      return pow(i, kMod - 2, kMod)

    @functools.lru_cache(None)
    def nCk(n: int, k: int) -> int:
      return fact(n) * inv(fact(k)) * inv(fact(n - k)) % kMod

    for freq in range(1, max(count.values()) + 1):
      numSubseqs = 1  # ""
      for charFreq in count.values():
        if charFreq >= freq:
          numSubseqs = numSubseqs * (1 + nCk(charFreq, freq)) % kMod
      ans += numSubseqs - 1  # Minus "".
      ans %= kMod

    return ans