Question

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 $C=5$。

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

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;
  }
};

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);
};

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 $C=5$。

class Solution:
  def countVowelPermutation(self, n: int) -> int:
    mod = 10**9 + 7

    def mul(a: List[List[int]], b: List[List[int]]) -> List[List[int]]:
      m, n = len(a), len(b[0])
      c = [[0] * n for _ in range(m)]
      for i in range(m):
        for j in range(n):
          for k in range(len(b)):
            c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % mod
      return c

    def pow(a: List[List[int]], n: int) -> List[int]:
      res = [[1] * len(a)]
      while n:
        if n & 1:
          res = mul(res, a)
        a = mul(a, a)
        n >>= 1
      return res

    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],
    ]
    res = pow(a, n - 1)
    return sum(map(sum, res)) % mod

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;
  }
}

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;
  }
};

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
}

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;
}

/**
 * @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;
}

方法三

import numpy as np


class Solution:
  def countVowelPermutation(self, n: int) -> int:
    mod = 10**9 + 7
    factor = np.mat(
      [
        (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.mat([(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

class Solution {
  public int countVowelPermutation(int n) {
    final int mod = 1000000007;
    long countA = 1, countE = 1, countI = 1, countO = 1, countU = 1;
    for (int length = 1; length < n; length++) {
      // Calculate the next counts for each vowel based on the previous counts
      long nextCountA = countE;
      long nextCountE = (countA + countI) % mod;
      long nextCountI = (countA + countE + countO + countU) % mod;
      long nextCountO = (countI + countU) % mod;
      long nextCountU = countA;
      // Update the counts with the newly calculated values for the next length
      countA = nextCountA;
      countE = nextCountE;
      countI = nextCountI;
      countO = nextCountO;
      countU = nextCountU;
    }
    // Calculate the total count of valid strings for length n
    long totalCount = (countA + countE + countI + countO + countU) % mod;
    return (int) totalCount;
  }
}