README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1600-1699/1641.Count%20Sorted%20Vowel%20Strings/README_EN.md	1519	Weekly Contest 213 Q2	Math Dynamic Programming Combinatorics

1641. Count Sorted Vowel Strings

中文文档

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

Solution 1

Python3

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)

Java

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;
    }
}

C++

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);
    }
};

Go

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)
}

Solution 2

Python3

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)

Java

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();
    }
}

C++

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);
    }
};

Go

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
}