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940. Distinct Subsequences II 👍

  • Time: $O(26n) = O(n)$
  • Space: $O(26) = O(1)$
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class Solution {
 public:
  int distinctSubseqII(string s) {
    constexpr int kMod = 1'000'000'007;
    // endsIn[i] := the number of subsequence that end in ('a' + i)
    vector<long> endsIn(26);

    for (const char c : s)
      endsIn[c - 'a'] = accumulate(endsIn.begin(), endsIn.end(), 1L) % kMod;

    return accumulate(endsIn.begin(), endsIn.end(), 0L) % kMod;
  }
};

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class Solution {
  public int distinctSubseqII(String s) {
    final int MOD = 1_000_000_007;
    // endsIn[i] := the number of subsequence that end in ('a' + i)
    long[] endsIn = new long[26];

    for (final char c : s.toCharArray())
      endsIn[c - 'a'] = (Arrays.stream(endsIn).sum() + 1) % MOD;

    return (int) (Arrays.stream(endsIn).sum() % MOD);
  }
}

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class Solution:
  def distinctSubseqII(self, s: str) -> int:
    MOD = 1_000_000_007
    # endsIn[i] := the number of subsequence that end in ('a' + i)
    endsIn = [0] * 26

    for c in s:
      endsIn[ord(c) - ord('a')] = (sum(endsIn) + 1) % MOD

    return sum(endsIn) % MOD