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1641. Count Sorted Vowel Strings

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

• Java
• C++
• Python
• Go

• 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);
      }
  };
  

• 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)
  
  

• 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)
  }
  

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