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1641. Count Sorted Vowel Strings

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

Thinking

A nondecreasing vowel string of length \(n\) is a weakly sorted selection from five letters. \(n\) is small, so search by “placed \(i\) letters, last vowel index \(j\)”.

\(dfs(i,j)\) returns \(1\) at \(i=n\), otherwise sums \(dfs(i+1,k)\) for \(k \ge j\), memoized to reuse states.

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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)

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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;
    }
}

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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);
    }
};

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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

Thinking

The 2-D state in Solution 1 collapses to counts ending at each vowel. The next length is a prefix sum of the previous counts.

Roll a length-\(5\) array, write prefix sums back, and sum it. Extra space is constant.

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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)

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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();
    }
}

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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);
    }
};

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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
}

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