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1220. Count Vowels Permutation

Description

Given an integer n, your task is to count how many strings of length n can be formed under the following rules:

  • Each character is a lower case vowel ('a', 'e', 'i', 'o', 'u')
  • Each vowel 'a' may only be followed by an 'e'.
  • Each vowel 'e' may only be followed by an 'a' or an 'i'.
  • Each vowel 'i' may not be followed by another 'i'.
  • Each vowel 'o' may only be followed by an 'i' or a 'u'.
  • Each vowel 'u' may only be followed by an 'a'.

Since the answer may be too large, return it modulo 10^9 + 7.

 

Example 1:

Input: n = 1
Output: 5
Explanation: All possible strings are: "a", "e", "i" , "o" and "u".

Example 2:

Input: n = 2
Output: 10
Explanation: All possible strings are: "ae", "ea", "ei", "ia", "ie", "io", "iu", "oi", "ou" and "ua".

Example 3: 

Input: n = 5
Output: 68

 

Constraints:

  • 1 <= n <= 2 * 10^4

Solutions

Solution 1: Dynamic Programming

Based on the problem description, we can list the possible subsequent vowels for each vowel:

a [e]
e [a|i]
i [a|e|o|u]
o [i|u]
u [a]

From this, we can deduce the possible preceding vowels for each vowel:

[e|i|u] a
[a|i]   e
[e|o]   i
[i] o
[i|o]   u

We define $f[i]$ as the number of strings of the current length ending with the $i$-th vowel. If the length is $1$, then $f[i]=1$.

When the length is greater than $1$, we define $g[i]$ as the number of strings of the current length ending with the $i$-th vowel. Then $g[i]$ can be derived from $f$, that is:

$$ g[i]= \begin{cases} f[1]+f[2]+f[4] & i=0 \ f[0]+f[2] & i=1 \ f[1]+f[3] & i=2 \ f[2] & i=3 \ f[2]+f[3] & i=4 \end{cases} $$

The final answer is $\sum_{i=0}^{4}f[i]$. Note that the answer may be very large, so we need to take the modulus of $10^9+7$.

The time complexity is $O(n)$, and the space complexity is $O(C)$. Here, $n$ is the length of the string, and $C$ is the number of vowels. In this problem, $C=5$.

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class Solution:
    def countVowelPermutation(self, n: int) -> int:
        f = [1] * 5
        mod = 10**9 + 7
        for _ in range(n - 1):
            g = [0] * 5
            g[0] = (f[1] + f[2] + f[4]) % mod
            g[1] = (f[0] + f[2]) % mod
            g[2] = (f[1] + f[3]) % mod
            g[3] = f[2]
            g[4] = (f[2] + f[3]) % mod
            f = g
        return sum(f) % mod

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class Solution {
    public int countVowelPermutation(int n) {
        long[] f = new long[5];
        Arrays.fill(f, 1);
        final int mod = (int) 1e9 + 7;
        for (int i = 1; i < n; ++i) {
            long[] g = new long[5];
            g[0] = (f[1] + f[2] + f[4]) % mod;
            g[1] = (f[0] + f[2]) % mod;
            g[2] = (f[1] + f[3]) % mod;
            g[3] = f[2];
            g[4] = (f[2] + f[3]) % mod;
            f = g;
        }
        long ans = 0;
        for (long x : f) {
            ans = (ans + x) % mod;
        }
        return (int) ans;
    }
}

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class Solution {
public:
    int countVowelPermutation(int n) {
        using ll = long long;
        vector<ll> f(5, 1);
        const int mod = 1e9 + 7;
        for (int i = 1; i < n; ++i) {
            vector<ll> g(5);
            g[0] = (f[1] + f[2] + f[4]) % mod;
            g[1] = (f[0] + f[2]) % mod;
            g[2] = (f[1] + f[3]) % mod;
            g[