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1220. Count Vowels Permutation

SourceWeekly Contest 157 Q4DifficultyHardRating1729

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

Thinking

Vowel strings of length \(n\) obey adjacency rules. \(n\) reaches \(2\times 10^4\), so we cannot list strings. It suffices to count strings by last vowel: each letter has a fixed set of legal predecessors.

An array \(f\) of length \(5\) stores counts by ending letter; we roll it \(n-1\) times by those predecessors. Only the previous length is needed. We reduce modulo \(10^9+7\).

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, \(C=5\).

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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

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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;
    }
}

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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;
    }
};

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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
}

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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);
}

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/**
 * @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);
};

Solution 2: Matrix Exponentiation to Accelerate Recursion

Thinking

Solution 1 spends \(O(n)\) constant-size updates. The five-state recurrence is a linear map. Encoded as a \(5\times 5\) matrix, exponentiation by squaring yields the \(n\)-th vector in \(O(\log n)\) multiplications, which helps for larger \(n\).

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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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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/**
 * @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;
}

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