1022EasyString

Is the Token Well Formed

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A token reads token, made up of English letters in either case, digits, and the marks '@', '#' and '$'.

The token is well formed when all of these hold:

it runs to at least three characters;
every character is a letter or a digit, so none of the three marks appears anywhere;
it holds at least one vowel, meaning 'a', 'e', 'i', 'o' or 'u' in either case;
it holds at least one consonant, meaning a letter that is not a vowel.

Return whether the token is well formed.

Examples

Example 1

Input: token = "aB3"
Output: true

Three characters, none of them a mark, with `a` as the vowel and `B` as the consonant. The digit is allowed and counts as neither.

Example 2

Input: token = "xyz"
Output: false

Long enough and free of marks, but every letter is a consonant, so there is no vowel.

Example 3

Input: token = "q@w"
Output: false

The mark in the middle is not allowed, whatever else the token holds.

Constraints

1 <= token.length <= 20
Every character of token is an English letter, a digit, or one of '@', '#' and '$'.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def is_valid(token: str) -> bool:

Java

public boolean isValid(String token)

The intended approach

Time O(n) over the token's length.Space O(1).

Check the length first, since a token shorter than three characters fails whatever it holds. Then one pass over the characters settles the rest. A digit is allowed but counts as neither a vowel nor a consonant, so pass over it. Anything that is not a letter must be one of the three marks, so reject the token there and then. A letter is a vowel or it is not, and each case sets its own flag, with case folded away by testing the letter against both spellings of the five vowels. The token is well formed when both flags ended up set. Nothing needs holding but the two flags.

Pythonentry point: is_valid

def is_valid(token):
    if len(token) < 3:
        return False
    vowels = "aeiouAEIOU"
    has_vowel = False
    has_consonant = False
    for ch in token:
        if ch.isdigit():
            continue
        if not ch.isalpha():
            return False
        if ch in vowels:
            has_vowel = True
        else:
            has_consonant = True
    return has_vowel and has_consonant

The obvious version

Spell the character groups out by hand instead of leaning on any built-in test: a string of the lowercase letters, a string of the uppercase ones, a string of the digits, and a string of the vowels. Fold each character to lowercase by looking it up in the uppercase string, then tally the vowels and consonants and reject anything that turns up in none of the groups.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_is_valid(token):
    lowers = "abcdefghijklmnopqrstuvwxyz"
    uppers = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
    digits = "0123456789"
    vowels = "aeiou"
    if len(token) < 3:
        return False
    seen_vowel = 0
    seen_consonant = 0
    for ch in token:
        if ch in digits:
            continue
        low = ch
        if ch in uppers:
            low = lowers[uppers.index(ch)]
        if low not in lowers:
            return False
        if low in vowels:
            seen_vowel += 1
        else:
            seen_consonant += 1
    return seen_vowel > 0 and seen_consonant > 0

How this problem was checked

Reference solution executed against 18 test cases — 18 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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