Count Distinct Vowels in String

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Given a string consisting of English alphabetic characters (both uppercase and lowercase), count the number of distinct vowels present. Vowels are 'a', 'e', 'i', 'o', 'u', and the count is case-insensitive (i.e., 'A' and 'a' are considered the same vowel).

Examples:

Input: "Hello"

Output: 2

Explanation: The string "Hello" contains vowels 'e' and 'o', which are distinct, so the count is 2.

Input: "AEiou"

Output: 5

Explanation: All five vowels ('a', 'e', 'i', 'o', 'u') are present, and since the count is case-insensitive, they are all distinct, resulting in a count of 5.

Input: "bcdfg"

Output: 0

Explanation: The string contains no vowels, so the count of distinct vowels is 0.

Constraints

The input string contains only English alphabetic characters (uppercase or lowercase).
1 <= length of string <= 1000

Editorial

Intuition:

We need to count distinct vowels case-insensitively. Using a set automatically handles uniqueness, and converting each character to lowercase ensures 'A' and 'a' are treated as the same vowel.

Approach:

Initialize an empty set to store distinct vowels.
Iterate through each character of the input string.
Convert the character to lowercase.
If the character is one of 'a', 'e', 'i', 'o', 'u', add it to the set.
After the loop, the set's size is the count of distinct vowels.

#include <iostream>
#include <unordered_set>
#include <cctype>
using namespace std;

int main() {
    string s;
    cin >> s;
    unordered_set<char> distinctVowels;
    for (char c : s) {
        c = tolower(c);
        if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
            distinctVowels.insert(c);
        }
    }
    cout << distinctVowels.size() << endl;
    return 0;
}
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        Set<Character> distinctVowels = new HashSet<>();
        for (char c : s.toCharArray()) {
            c = Character.toLowerCase(c);
            if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
                distinctVowels.add(c);
            }
        }
        System.out.println(distinctVowels.size());
    }
}
s = input().strip()
vowels = set()
for c in s:
    c = c.lower()
    if c in ['a','e','i','o','u']:
        vowels.add(c)
print(len(vowels))
const s = require('fs').readFileSync(0, 'utf8').trim();

let distinctVowels = new Set();
for (let c of s) {
    c = c.toLowerCase();
    if (c === 'a' || c === 'e' || c === 'i' || c === 'o' || c === 'u') {
        distinctVowels.add(c);
    }
}
console.log(distinctVowels.size);

Complexity Analysis:

Time Complexity: O(n), where n is the string length, as we process each character once and set operations are O(1) average.

Space Complexity: O(1), since the set stores at most 5 vowels (constant space).

Notes

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