Problem

Given a string s, return three counts: the number of vowels (a e i o u, either case), the number of consonants (all other English letters), and the number of spaces. Digits and punctuation are ignored.

Input / Output

• Input: string s.
• Output: [vowels, consonants, spaces].

Examples

Example 1 Input: s = "Hello World" Output: [3, 7, 1]

Example 2 Input: s = "AEIOU xyz" Output: [5, 3, 1]

Example 3 Input: s = "123 !!" Output: [0, 0, 1]

Constraints

• 0 <= s.length <= 10^5

Approach

Single pass; classify each character:

1. If it is a space → spaces++.
2. Else if it is a letter → lowercase it; if it is in {a,e,i,o,u} → vowels++, else consonants++.
3. Otherwise ignore.

Use a set for the vowel test (O(1)) rather than a chain of or comparisons. Note that y counts as a consonant here.

Complexity

• Time: O(n)
• Space: O(1)

Reference solution (Python)

def count_vcs(s):
    vowels = consonants = spaces = 0
    V = set("aeiou")
    for c in s:
        if c == ' ':
            spaces += 1
        elif c.isalpha():
            if c.lower() in V:
                vowels += 1
            else:
                consonants += 1
    return [vowels, consonants, spaces]

Test cases

Input	Expected
"Hello World"	[3,7,1]
"AEIOU xyz"	[5,3,1]
"123 !!"	[0,0,1]
""	[0,0,0]