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
SOLUTION:
class Solution:
def cvow(self, n):
if n == 1:
return [5, 4, 3, 2, 1]
curr = self.cvow(n - 1)
op = [curr[-1]]
for i in range(3, -1, -1):
op.append(op[-1] + curr[i])
return op[::-1]
def countVowelStrings(self, n: int) -> int:
return self.cvow(n)[0]
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