LeetCode 1220. Count Vowels Permutation Shell Solution | Explanation + Code

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 ∑_{i=0}⁴f[i]. Note that the answer may be very large, so we need to take the modulus of 10⁹+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.

PythonJavaC++GoTypeScriptJavaScript

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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); }(code-box)

/** * @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); };(code-box)

Solution 2: Matrix Exponentiation to Accelerate Recursion

The time complexity is O(C³ × log n), and the space complexity is O(C²). Here, C is the number of vowels. In this problem, C=5.

PythonJavaC++GoTypeScriptJavaScript

import numpy as np class Solution: def countVowelPermutation(self, n: int) -> int: mod = 10**9 + 7 factor = np.asmatrix( [ (0, 1, 0, 0, 0), (1, 0, 1, 0, 0), (1, 1, 0, 1, 1), (0, 0, 1, 0, 1), (1, 0, 0, 0, 0), ], np.dtype("O"), ) res = np.asmatrix([(1, 1, 1, 1, 1)], np.dtype("O")) n -= 1 while n: if n & 1: res = res * factor % mod factor = factor * factor % mod n >>= 1 return res.sum() % mod(code-box)

class Solution { private final int mod = (int) 1e9 + 7; public int countVowelPermutation(int n) { long[][] a = {{0, 1, 0, 0, 0}, {1, 0, 1, 0, 0}, {1, 1, 0, 1, 1}, {0, 0, 1, 0, 1}, {1, 0, 0, 0, 0}}; long[][] res = pow(a, n - 1); long ans = 0; for (long x : res[0]) { ans = (ans + x) % mod; } return (int) ans; } private long[][] mul(long[][] a, long[][] b) { int m = a.length, n = b[0].length; long[][] c = new long[m][n]; for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { for (int k = 0; k < b.length; ++k) { c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % mod; } } } return c; } private long[][] pow(long[][] a, int n) { long[][] res = new long[1][a.length]; Arrays.fill(res[0], 1); while (n > 0) { if ((n & 1) == 1) { res = mul(res, a); } a = mul(a, a); n >>= 1; } return res; } }(code-box)

class Solution { public: int countVowelPermutation(int n) { vector<vector<ll>> a = { {0, 1, 0, 0, 0}, {1, 0, 1, 0, 0}, {1, 1, 0, 1, 1}, {0, 0, 1, 0, 1}, {1, 0, 0, 0, 0}}; vector<vector<ll>> res = pow(a, n - 1); return accumulate(res[0].begin(), res[0].end(), 0LL) % mod; } private: using ll = long long; const int mod = 1e9 + 7; vector<vector<ll>> mul(vector<vector<ll>>& a, vector<vector<ll>>& b) { int m = a.size(), n = b[0].size(); vector<vector<ll>> c(m, vector<ll>(n)); for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { for (int k = 0; k < b.size(); ++k) { c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % mod; } } } return c; } vector<vector<ll>> pow(vector<vector<ll>>& a, int n) { vector<vector<ll>> res; res.push_back({1, 1, 1, 1, 1}); while (n) { if (n & 1) { res = mul(res, a); } a = mul(a, a); n >>= 1; } return res; } };(code-box)

const mod = 1e9 + 7 func countVowelPermutation(n int) (ans int) { a := [][]int{ {0, 1, 0, 0, 0}, {1, 0, 1, 0, 0}, {1, 1, 0, 1, 1}, {0, 0, 1, 0, 1}, {1, 0, 0, 0, 0}} res := pow(a, n-1) for _, x := range res[0] { ans = (ans + x) % mod } return } func mul(a, b [][]int) [][]int { m, n := len(a), len(b[0]) c := make([][]int, m) for i := range c { c[i] = make([]int, n) } for i := 0; i < m; i++ { for j := 0; j < n; j++ { for k := 0; k < len(b); k++ { c[i][j] = (c[i][j] + a[i][k]*b[k][j]) % mod } } } return c } func pow(a [][]int, n int) [][]int { res := [][]int{{1, 1, 1, 1, 1}} for n > 0 { if n&1 == 1 { res = mul(res, a) } a = mul(a, a) n >>= 1 } return res }(code-box)

const mod = 1e9 + 7; function countVowelPermutation(n: number): number { const a: number[][] = [ [0, 1, 0, 0, 0], [1, 0, 1, 0, 0], [1, 1, 0, 1, 1], [0, 0, 1, 0, 1], [1, 0, 0, 0, 0], ]; const res = pow(a, n - 1); return res[0].reduce((a, b) => (a + b) % mod); } function mul(a: number[][], b: number[][]): number[][] { const [m, n] = [a.length, b[0].length]; const c = Array.from({ length: m }, () => Array.from({ length: n }, () => 0)); for (let i = 0; i < m; ++i) { for (let j = 0; j < n; ++j) { for (let k = 0; k < b.length; ++k) { c[i][j] = (c[i][j] + Number((BigInt(a[i][k]) * BigInt(b[k][j])) % BigInt(mod))) % mod; } } } return c; } function pow(a: number[][], n: number): number[][] { let res: number[][] = [[1, 1, 1, 1, 1]]; while (n) { if (n & 1) { res = mul(res, a); } a = mul(a, a); n >>>= 1; } return res; }(code-box)

/** * @param {number} n * @return {number} */ const mod = 1e9 + 7; var countVowelPermutation = function (n) { const a = [ [0, 1, 0, 0, 0], [1, 0, 1, 0, 0], [1, 1, 0, 1, 1], [0, 0, 1, 0, 1], [1, 0, 0, 0, 0], ]; const res = pow(a, n - 1); return res[0].reduce((a, b) => (a + b) % mod); }; function mul(a, b) { const [m, n] = [a.length, b[0].length]; const c = Array.from({ length: m }, () => Array.from({ length: n }, () => 0)); for (let i = 0; i < m; ++i) { for (let j = 0; j < n; ++j) { for (let k = 0; k < b.length; ++k) { c[i][j] = (c[i][j] + Number((BigInt(a[i][k]) * BigInt(b[k][j])) % BigInt(mod))) % mod; } } } return c; } function pow(a, n) { let res = [[1, 1, 1, 1, 1]]; while (n) { if (n & 1) { res = mul(res, a); } a = mul(a, a); n >>>= 1; } return res; }(code-box)