Formatted question description: https://leetcode.ca/all/2062.html

2062. Count Vowel Substrings of a String (Easy)

A substring is a contiguous (non-empty) sequence of characters within a string.

A vowel substring is a substring that only consists of vowels ('a', 'e', 'i', 'o', and 'u') and has all five vowels present in it.

Given a string word, return the number of vowel substrings in word.

Example 1:

Input: word = "aeiouu"
Output: 2
Explanation: The vowel substrings of word are as follows (underlined):
- "aeiouu"
- "aeiouu"

Example 2:

Input: word = "unicornarihan"
Output: 0
Explanation: Not all 5 vowels are present, so there are no vowel substrings.

Example 3:

Input: word = "cuaieuouac"
Output: 7
Explanation: The vowel substrings of word are as follows (underlined):
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"

Example 4:

Input: word = "bbaeixoubb"
Output: 0
Explanation: The only substrings that contain all five vowels also contain consonants, so there are no vowel substrings.

Constraints:

1 <= word.length <= 100
word consists of lowercase English letters only.

Similar Questions:

Solution 1. Brute Force

// OJ: https://leetcode.com/problems/count-vowel-substrings-of-a-string/
// Time: O(N^2)
// Space: O(1)
class Solution {
    bool isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    };
public:
    int countVowelSubstrings(string s) {
        int ans = 0, N = s.size();
        unordered_map<char, int> cnt;
        for (int i = 0; i < N; ++i) {
            cnt.clear();
            for (int j = i; j < N && isVowel(s[j]); ++j) {
                cnt[s[j]]++;
                if (cnt.size() == 5) ++ans;
            }
        }
        return ans;
    }
};

class Solution:
    def countVowelSubstrings(self, word: str) -> int:
        n = len(word)
        s = set('aeiou')
        return sum(set(word[i:j]) == s for i in range(n) for j in range(i + 1, n + 1))

############

# 2062. Count Vowel Substrings of a String
# https://leetcode.com/problems/count-vowel-substrings-of-a-string/

class Solution:
    def countVowelSubstrings(self, word: str) -> int:
        res = 0
        
        n = len(word)
        
        for i in range(n):
            A = [0, 0, 0, 0, 0]
            for j in range(i, n):
                if word[j] == 'a':
                    A[0] += 1
                elif word[j] == 'e':
                    A[1] += 1
                elif word[j] == 'i':
                    A[2] += 1
                elif word[j] == 'o':
                    A[3] += 1
                elif word[j] == 'u':
                    A[4] += 1
                else:
                    break
                
                if all(x > 0 for x in A):
                    res += 1

        return res

class Solution {
    public int countVowelSubstrings(String word) {
        int n = word.length();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            Set<Character> t = new HashSet<>();
            for (int j = i; j < n; ++j) {
                char c = word.charAt(j);
                if (!isVowel(c)) {
                    break;
                }
                t.add(c);
                if (t.size() == 5) {
                    ++ans;
                }
            }
        }
        return ans;
    }

    private boolean isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    }
}

func countVowelSubstrings(word string) int {
	ans, n := 0, len(word)
	for i := range word {
		t := map[byte]bool{}
		for j := i; j < n; j++ {
			c := word[j]
			if !(c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
				break
			}
			t[c] = true
			if len(t) == 5 {
				ans++
			}
		}
	}
	return ans
}

function countVowelSubstrings(word: string): number {
    let ans = 0;
    const n = word.length;
    for (let i = 0; i < n; ++i) {
        const t = new Set<string>();
        for (let j = i; j < n; ++j) {
            const c = word[j];
            if (
                !(c === 'a' || c === 'e' || c === 'i' || c === 'o' || c === 'u')
            ) {
                break;
            }
            t.add(c);
            if (t.size === 5) {
                ans++;
            }
        }
    }
    return ans;
}

Solution 2. Sliding Window

Check out “C++ Maximum Sliding Window Cheatsheet Template!”

Function atMost(s, goal) returns the number of substrings that has at most goal number of unique vowels. The answer is atMost(s, 5) - atMost(s, 4).

// OJ: https://leetcode.com/problems/count-vowel-substrings-of-a-string/
// Time: O(N)
// Space: O(1)
class Solution {
    bool isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    };
    int atMost(string &s, int goal) {
        int ans = 0, i = 0, j = 0, N = s.size();
        unordered_map<char, int> cnt;
        for (; j < N; ++j) {
            if (!isVowel(s[j])) {
                i = j + 1;
                cnt.clear();
                continue;
            }
            cnt[s[j]]++;
            for (; cnt.size() > goal; ++i) {
                if (--cnt[s[i]] == 0) cnt.erase(s[i]);
            }
            ans += j - i + 1; // this window [i, j] is the maximum window ending at `s[j]` that has at most `goal` number of unique vowels.
        }
        return ans;
    }
public:
    int countVowelSubstrings(string s) {
        return atMost(s, 5) - atMost(s, 4);
    }
};

class Solution:
    def countVowelSubstrings(self, word: str) -> int:
        n = len(word)
        s = set('aeiou')
        return sum(set(word[i:j]) == s for i in range(n) for j in range(i + 1, n + 1))

############

# 2062. Count Vowel Substrings of a String
# https://leetcode.com/problems/count-vowel-substrings-of-a-string/

class Solution:
    def countVowelSubstrings(self, word: str) -> int:
        res = 0
        
        n = len(word)
        
        for i in range(n):
            A = [0, 0, 0, 0, 0]
            for j in range(i, n):
                if word[j] == 'a':
                    A[0] += 1
                elif word[j] == 'e':
                    A[1] += 1
                elif word[j] == 'i':
                    A[2] += 1
                elif word[j] == 'o':
                    A[3] += 1
                elif word[j] == 'u':
                    A[4] += 1
                else:
                    break
                
                if all(x > 0 for x in A):
                    res += 1

        return res

class Solution {
    public int countVowelSubstrings(String word) {
        int n = word.length();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            Set<Character> t = new HashSet<>();
            for (int j = i; j < n; ++j) {
                char c = word.charAt(j);
                if (!isVowel(c)) {
                    break;
                }
                t.add(c);
                if (t.size() == 5) {
                    ++ans;
                }
            }
        }
        return ans;
    }

    private boolean isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    }
}

func countVowelSubstrings(word string) int {
	ans, n := 0, len(word)
	for i := range word {
		t := map[byte]bool{}
		for j := i; j < n; j++ {
			c := word[j]
			if !(c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
				break
			}
			t[c] = true
			if len(t) == 5 {
				ans++
			}
		}
	}
	return ans
}

function countVowelSubstrings(word: string): number {
    let ans = 0;
    const n = word.length;
    for (let i = 0; i < n; ++i) {
        const t = new Set<string>();
        for (let j = i; j < n; ++j) {
            const c = word[j];
            if (
                !(c === 'a' || c === 'e' || c === 'i' || c === 'o' || c === 'u')
            ) {
                break;
            }
            t.add(c);
            if (t.size === 5) {
                ans++;
            }
        }
    }
    return ans;
}

Discuss

https://leetcode.com/problems/count-vowel-substrings-of-a-string/discuss/1563765/C%2B%2B-O(N)-Time-Sliding-Window

2062 - Count Vowel Substrings of a String

2062. Count Vowel Substrings of a String (Easy)

Solution 1. Brute Force

Solution 2. Sliding Window

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2062. Count Vowel Substrings of a String (Easy)

Solution 1. Brute Force

Solution 2. Sliding Window

Discuss

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All Solutions