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
a e i o u
1 1 1 1 1 n=1
5 4 3 2 1 n=2
15 10 6 3 1 n=3
... n=...class Solution:
def countVowelStrings(self, n: int) -> int:
cnt = [1] * 5
for i in range(2, n + 1):
for j in range(3, -1, -1):
cnt[j] += cnt[j + 1]
return sum(cnt)class Solution {
public int countVowelStrings(int n) {
int[] cnt = new int[5];
Arrays.fill(cnt, 1);
for (int i = 2; i <= n; ++i) {
for (int j = 3; j >= 0; --j) {
cnt[j] += cnt[j + 1];
}
}
return Arrays.stream(cnt).sum();
}
}class Solution {
public:
int countVowelStrings(int n) {
vector<int> cnt(5, 1);
for (int i = 2; i <= n; ++i)
for (int j = 3; j >= 0; --j)
cnt[j] += cnt[j + 1];
return accumulate(cnt.begin(), cnt.end(), 0);
}
};func countVowelStrings(n int) int {
cnt := make([]int, 5)
for i := range cnt {
cnt[i] = 1
}
for i := 2; i <= n; i++ {
for j := 3; j >= 0; j-- {
cnt[j] += cnt[j+1]
}
}
ans := 0
for _, v := range cnt {
ans += v
}
return ans
}