Problem
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 (Java)
class Solution {
public int countVowelStrings(int n) {
if (n == 1) {
return 5;
}
int[] arr = new int[] {1, 1, 1, 1, 1};
int sum = 5;
for (int i = 2; i <= n; i++) {
int[] copy = new int[5];
for (int j = 0; j < arr.length; j++) {
if (j == 0) {
copy[j] = sum;
} else {
copy[j] = copy[j - 1] - arr[j - 1];
}
}
arr = Arrays.copyOf(copy, 5);
sum = 0;
for (int k : arr) {
sum += k;
}
}
return sum;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).