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2539. Count the Number of Good Subsequences
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 is24.
There are five subsequences which are not good: "aabb", "aabb", "aabb", "aabb", and the empty subsequence. Hence, the number of good subsequences is24-5 = 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
-
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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int N = 10001; int MOD = 1e9 + 7; long f[10001]; long g[10001]; 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; } int init = []() { 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); } return 0; }(); int comb(int n, int k) { return (f[n] * g[k] % MOD) * g[n - k] % MOD; } class Solution { public: int countGoodSubsequences(string s) { int cnt[26]{}; int mx = 1; for (char& c : s) { mx = max(mx, ++cnt[c - '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 ans; } };
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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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const n = 1e4 + 1 const mod = 1e9 + 7 var f = make([]int, n) var g = make([]int, n) func qmi(a, k, p int) int { res := 1 for k != 0 { if k&1 == 1 { res = res * a % p } k >>= 1 a = a * a % p } return res } func init() { f[0], g[0] = 1, 1 for i := 1; i < n; i++ { f[i] = f[i-1] * i % mod g[i] = qmi(f[i], mod-2, mod) } } func comb(n, k int) int { return (f[n] * g[k] % mod) * g[n-k] % mod } func countGoodSubsequences(s string) (ans int) { cnt := [26]int{} mx := 1 for _, c := range s { cnt[c-'a']++ mx = max(mx, cnt[c-'a']) } for i := 1; i <= mx; i++ { x := 1 for _, v := range cnt { if v >= i { x = (x * (comb(v, i) + 1)) % mod } } ans = (ans + x - 1) % mod } return }