1+ /*
2+ Count Vowels Permutation
3+ ========================
4+
5+ Given an integer n, your task is to count how many strings of length n can be formed under the following rules:
6+
7+ Each character is a lower case vowel ('a', 'e', 'i', 'o', 'u')
8+ Each vowel 'a' may only be followed by an 'e'.
9+ Each vowel 'e' may only be followed by an 'a' or an 'i'.
10+ Each vowel 'i' may not be followed by another 'i'.
11+ Each vowel 'o' may only be followed by an 'i' or a 'u'.
12+ Each vowel 'u' may only be followed by an 'a'.
13+ Since the answer may be too large, return it modulo 10^9 + 7.
14+
15+ Example 1:
16+ Input: n = 1
17+ Output: 5
18+ Explanation: All possible strings are: "a", "e", "i" , "o" and "u".
19+
20+ Example 2:
21+ Input: n = 2
22+ Output: 10
23+ Explanation: All possible strings are: "ae", "ea", "ei", "ia", "ie", "io", "iu", "oi", "ou" and "ua".
24+
25+ Example 3:
26+ Input: n = 5
27+ Output: 68
28+
29+ Constraints:
30+ 1 <= n <= 2 * 10^4
31+
32+ Hint #1
33+ Use dynamic programming.
34+
35+ Hint #2
36+ Let dp[i][j] be the number of strings of length i that ends with the j-th vowel.
37+
38+ Hint #3
39+ Deduce the recurrence from the given relations between vowels.
40+ */
41+
42+ class Solution
43+ {
44+ public:
45+ int M = 1e9 + 7 ;
46+
47+ int countVowelPermutation (int n)
48+ {
49+ long long a = 1 , e = 1 , i = 1 , o = 1 , u = 1 ;
50+
51+ for (int j = 1 ; j < n; ++j)
52+ {
53+ long long A = 0 , E = 0 , I = 0 , O = 0 , U = 0 ;
54+ A = e + i + u;
55+ E = a + i;
56+ I = e + o;
57+ O = i;
58+ U = i + o;
59+
60+ a = A % M;
61+ e = E % M;
62+ i = I % M;
63+ o = O % M;
64+ u = U % M;
65+ }
66+
67+ return (a % M + e % M + i % M + o % M + u % M) % M;
68+ }
69+ };
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