LeetCode //C - 1220. Count Vowels Permutation

1220. Count Vowels Permutation

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

From: LeetCode

Link: 1220. Count Vowels Permutation

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Solution:

Ideas:

keep counts of strings ending with each vowel, then update by reverse rules.

Code:

int countVowelPermutation(int n) {
    const long MOD = 1000000007;

    long a = 1, e = 1, i = 1, o = 1, u = 1;

    for (int len = 2; len <= n; len++) {
        long na = (e + i + u) % MOD;      // previous e/i/u can go to a
        long ne = (a + i) % MOD;          // previous a/i can go to e
        long ni = (e + o) % MOD;          // previous e/o can go to i
        long no = i % MOD;                // previous i can go to o
        long nu = (i + o) % MOD;          // previous i/o can go to u

        a = na;
        e = ne;
        i = ni;
        o = no;
        u = nu;
    }

    return (int)((a + e + i + o + u) % MOD);
}