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2539. Count the Number of Good Subsequences πŸ”’

Description

A subsequence of a string is good if it is not empty and the frequency of each one of its characters is the same.

Given a string s, return the number of good subsequences of s. Since the answer may be too large, return it modulo 109 + 7.

A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

 

Example 1:

Input: s = "aabb"
Output: 11
Explanation: The total number of subsequences is 24. There are five subsequences which are not good: "aabb", "aabb", "aabb", "aabb", and the empty subsequence. Hence, the number of good subsequences is 24-5 = 11.

Example 2:

Input: s = "leet"
Output: 12
Explanation: There are four subsequences which are not good: "leet", "leet", "leet", and the empty subsequence. Hence, the number of good subsequences is 24-4 = 12.

Example 3:

Input: s = "abcd"
Output: 15
Explanation: All of the non-empty subsequences are good subsequences. Hence, the number of good subsequences is 24-1 = 15.

 

Constraints:

  • 1 <= s.length <= 104
  • s consists of only lowercase English letters.

Solutions

Solution 1

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N = 10001
MOD = 10**9 + 7
f = [1] * N
g = [1] * N
for i in range(1, N):
    f[i] = f[i - 1] * i % MOD
    g[i] = pow(f[i], MOD - 2, MOD)


def comb(n, k):
    return f[n] * g[k] * g[n - k] % MOD


class Solution:
    def countGoodSubsequences(self, s: str) -> int:
        cnt = Counter(s)
        ans = 0
        for i in range(1, max(cnt.values()) + 1):
            x = 1
            for v in cnt.values():
                if v >= i:
                    x = x * (comb(v, i) + 1) % MOD
            ans = (ans + x - 1) % MOD
        return ans

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class Solution {
    private static final int N = 10001;
    private static final int MOD = (int) 1e9 + 7;
    private static final long[] F = new long[N];
    private static final long[] G = new long[N];

    static {
        F[0] = 1;
        G[0] = 1;
        for (int i = 1; i < N; ++i) {
            F[i] = F[i - 1] * i % MOD;
            G[i] = qmi(F[i], MOD - 2, MOD);
        }
    }

    public static long qmi(long a, long k, long p) {
        long res = 1;
        while (k != 0) {
            if ((k & 1) == 1) {
                res = res * a % p;
            }
            k >>= 1;
            a = a * a % p;
        }
        return res;
    }

    public static long comb(int n, int k) {
        return (F[n] * G[k] % MOD) * G[n - k] % MOD;
    }

    public int countGoodSubsequences(String s) {
        int[] cnt = new int[26];
        int mx = 1;
        for (int i = 0; i < s.length(); ++i) {
            mx = Math.max(mx, ++cnt[s.charAt(i) - 'a']);
        }
        long ans = 0;
        for (int i = 1; i <= mx; ++i) {
            long x = 1;
            for (int j = 0; j < 26; ++j) {
                if (cnt[j] >= i) {
                    x = x * (comb(cnt[j], i) + 1) % MOD;
                }
            }
            ans = (ans + x - 1) % MOD;
        }
        return (int) ans;
    }
}

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