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困难
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第 157 场周赛 Q4
动态规划

1220. 统计元音字母序列的数目

English Version

题目描述

给你一个整数 n，请你帮忙统计一下我们可以按下述规则形成多少个长度为 n 的字符串：

  • 字符串中的每个字符都应当是小写元音字母（'a', 'e', 'i', 'o', 'u'）
  • 每个元音 'a' 后面都只能跟着 'e'
  • 每个元音 'e' 后面只能跟着 'a' 或者是 'i'
  • 每个元音 'i' 后面 不能 再跟着另一个 'i'
  • 每个元音 'o' 后面只能跟着 'i' 或者是 'u'
  • 每个元音 'u' 后面只能跟着 'a'

由于答案可能会很大，所以请你返回 模 10^9 + 7 之后的结果。

 

示例 1：

输入：n = 1
输出：5
解释：所有可能的字符串分别是："a", "e", "i" , "o" 和 "u"。

示例 2：

输入：n = 2
输出：10
解释：所有可能的字符串分别是："ae", "ea", "ei", "ia", "ie", "io", "iu", "oi", "ou" 和 "ua"。

示例 3：

输入：n = 5
输出：68

 

提示：

  • 1 <= n <= 2 * 10^4

解法

方法一：动态规划

根据题目描述，我们先列出每个元音字母的后一个字母：

a [e]
e [a|i]
i [a|e|o|u]
o [i|u]
u [a]

那么我们可以推出每个元音字母的前一个字母：

[e|i|u]	a
[a|i]	e
[e|o]	i
[i]	o
[i|o]	u

我们定义 $f[i]$ 表示当前长度以第 $i$ 个元音字母结尾的字符串的个数，如果长度为 $1$，那么 $f[i]=1$。

当长度大于 $1$ 时，我们定义 $g[i]$ 表示当前长度以第 $i$ 个元音字母结尾的字符串的个数，那么 $g[i]$ 可以由 $f$ 转移而来，即：

$$ g[i]= \begin{cases} f[1]+f[2]+f[4] & i=0 \\ f[0]+f[2] & i=1 \\ f[1]+f[3] & i=2 \\ f[2] & i=3 \\ f[2]+f[3] & i=4 \end{cases} $$

最终答案为 $\sum_{i=0}^{4}f[i]$。注意由于答案可能会很大，所以需要对 $10^9+7$ 取模。

时间复杂度 $O(n)$，空间复杂度 $O(C)$。其中 $n$ 是字符串的长度，而 $C$ 是元音字母的个数。本题中 $C=5$。

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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);
}

JavaScript

/**
 * @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);
};

方法二：矩阵快速幂加速递推

时间复杂度 $O(C^3 \times \log n)$，空间复杂度 $O(C^2)$，其中 $C$ 是元音字母的个数，本题中 $C=5$。

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

JavaScript

/**
 * @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;
}