Question

1641. Count Sorted Vowel Strings

中文文档

Description

Given an integer n, return _the number of strings of length _n_ that consist only of vowels (_a_, _e_, _i_, _o_, _u_) and are lexicographically sorted._

A string s is lexicographically sorted if for all valid i, s[i] is the same as or comes before s[i+1] in the alphabet.

 

Example 1:

Input: n = 1
Output: 5
Explanation: The 5 sorted strings that consist of vowels only are ["a","e","i","o","u"].

Example 2:

Input: n = 2
Output: 15
Explanation: The 15 sorted strings that consist of vowels only are
["aa","ae","ai","ao","au","ee","ei","eo","eu","ii","io","iu","oo","ou","uu"].
Note that "ea" is not a valid string since 'e' comes after 'a' in the alphabet.

Example 3:

Input: n = 33
Output: 66045

 

Constraints:

• 1 <= n <= 50 

Solutions

Solution 1

class Solution:
  def countVowelStrings(self, n: int) -> int:
    @cache
    def dfs(i, j):
      return 1 if i >= n else sum(dfs(i + 1, k) for k in range(j, 5))

    return dfs(0, 0)

class Solution {
  private Integer[][] f;
  private int n;

  public int countVowelStrings(int n) {
    this.n = n;
    f = new Integer[n][5];
    return dfs(0, 0);
  }

  private int dfs(int i, int j) {
    if (i >= n) {
      return 1;
    }
    if (f[i][j] != null) {
      return f[i][j];
    }
    int ans = 0;
    for (int k = j; k < 5; ++k) {
      ans += dfs(i + 1, k);
    }
    return f[i][j] = ans;
  }
}

class Solution {
public:
  int countVowelStrings(int n) {
    int f[n][5];
    memset(f, 0, sizeof f);
    function<int(int, int)> dfs = [&](int i, int j) {
      if (i >= n) {
        return 1;
      }
      if (f[i][j]) {
        return f[i][j];
      }
      int ans = 0;
      for (int k = j; k < 5; ++k) {
        ans += dfs(i + 1, k);
      }
      return f[i][j] = ans;
    };
    return dfs(0, 0);
  }
};

func countVowelStrings(n int) int {
  f := make([][5]int, n)
  var dfs func(i, j int) int
  dfs = func(i, j int) int {
    if i >= n {
      return 1
    }
    if f[i][j] != 0 {
      return f[i][j]
    }
    ans := 0
    for k := j; k < 5; k++ {
      ans += dfs(i+1, k)
    }
    f[i][j] = ans
    return ans
  }
  return dfs(0, 0)
}

Solution 2

class Solution:
  def countVowelStrings(self, n: int) -> int:
    f = [1] * 5
    for _ in range(n - 1):
      s = 0
      for j in range(5):
        s += f[j]
        f[j] = s
    return sum(f)

class Solution {
  public int countVowelStrings(int n) {
    int[] f = {1, 1, 1, 1, 1};
    for (int i = 0; i < n - 1; ++i) {
      int s = 0;
      for (int j = 0; j < 5; ++j) {
        s += f[j];
        f[j] = s;
      }
    }
    return Arrays.stream(f).sum();
  }
}

class Solution {
public:
  int countVowelStrings(int n) {
    int f[5] = {1, 1, 1, 1, 1};
    for (int i = 0; i < n - 1; ++i) {
      int s = 0;
      for (int j = 0; j < 5; ++j) {
        s += f[j];
        f[j] = s;
      }
    }
    return accumulate(f, f + 5, 0);
  }
};

func countVowelStrings(n int) (ans int) {
  f := [5]int{1, 1, 1, 1, 1}
  for i := 0; i < n-1; i++ {
    s := 0
    for j := 0; j < 5; j++ {
      s += f[j]
      f[j] = s
    }
  }
  for _, v := range f {
    ans += v
  }
  return
}