# 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

Solution:

Easy DP, just count the number starting with any letter following the rule.

int mod = 1e9+7;
class Solution {
public:
int countVowelPermutation(int n) {
vector<vector<int>> dp(n+1,vector<int>(5,0));
dp[1] = {1,1,1,1,1};
for(int i =2;i<=n; i++ ){
dp[i][0] = dp[i-1][1] % mod;
dp[i][1] = (dp[i-1][0] + dp[i-1][2]) % mod;
dp[i][2] = ((dp[i-1][0] + dp[i-1][1]) % mod + (dp[i-1][3] + dp[i-1][4]) % mod) % mod;
dp[i][3] = (dp[i-1][2] + dp[i-1][4]) % mod;
dp[i][4] = dp[i-1][0] % mod;
}
return (((dp[n][0]+dp[n][1])%mod +(dp[n][2]+dp[n][3])%mod)%mod+dp[n][4])%mod;
}
};