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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, $C=5$.
Solution 2: Matrix Exponentiation to Accelerate Recursion
The time complexity is $O(C^3 \times \log n)$, and the space complexity is $O(C^2)$. Here, $C$ is the number of vowels. In this problem, $C=5$.
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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; } } -
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[3] = f[2]; g[4] = (f[2] + f[3]) % mod; f = move(g); } return accumulate(f.begin(), f.end(), 0LL) % mod; } }; -
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 -
func countVowelPermutation(n int) (ans int) { const mod int = 1e9 + 7 f := make([]int, 5) for i := range f { f[i] = 1 } for i := 1; i < n; i++ { g := make([]int, 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] % mod g[4] = (f[2] + f[3]) % mod f = g } for _, x := range f { ans = (ans + x) % mod } return } -
function countVowelPermutation(n: number): number { const f: number[] = Array(5).fill(1); const mod = 1e9 + 7; for (let i = 1; i < n; ++i) { const g: number[] = Array(5).fill(0); 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.splice(0, 5, ...g); } return f.reduce((a, b) => (a + b) % mod); } -
/** * @param {number} n * @return {number} */ var countVowelPermutation = function (n) { const mod = 1e9 + 7; const f = Array(5).fill(1); for (let i = 1; i < n; ++i) { const g = Array(5).fill(0); 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.splice(0, 5, ...g); } return f.reduce((a, b) => (a + b) % mod); };