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:
Solutions
Solution 1
PythonJavaC++Go
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)(code-box)
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;
}
}(code-box)
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);
}
};(code-box)
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)
}(code-box)
Solution 2
PythonJavaC++Go
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)(code-box)
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();
}
}(code-box)
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);
}
};(code-box)
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
}(code-box)
