Is it a strong word?

Question

They say that hate is a strong word. I wanted to find out why, so I had a good look at the word.

I noticed that every consonant had a vowel after it. That made it look quite strong to me, so I decided that that's what makes a word strong.

I want to find more strong words, so I'll need a program for it!

Finding strong words

Strong words are words where every consonant (letters in the set BCDFGHJKLMNPQRSTVWXZ) is followed by a vowel (letters in the set AEIOUY). That's it. Nothing else matters.

If the word starts with a vowel, you don't have to worry about any of the letters before the first consonant. If the word has no consonants in it at all, it's automatically a strong word!

Some examples of strong words are agate, hate and you. agate is still a strong word because although it starts with a vowel, every consonant is still followed by a vowel. you is a strong word because it has no consonants.

There is no restriction on length for strong words.

The challenge

Write a program or function that takes a non-empty string as input, and outputs a truthy value if it is a strong word or a falsy value if it is not.

Clarifications

• You may decide to take the input in either lowercase or uppercase. Specify which in your answer.
• Words will not contain punctuation of any kind. They will only contain plain letters in the set ABCDEFGHIJKLMNOPQRSTUVWXYZ.
• Instead of truthy and falsy values, you may choose two distinct and consistent values to return for true and false. If you do this, specify the values you have picked in your answer.
  • You may alternatively output a falsy value for a strong word and a truthy one for a non-strong word.

Test cases

Input      -> Output
hate       -> true
love       -> true
popularize -> true
academy    -> true
you        -> true
mouse      -> true
acorn      -> false
nut        -> false
ah         -> false
strong     -> false
false      -> false
parakeet   -> false

Scoring

Since this is code-golf, the answer with the least bytes wins!

Answer 1

JavaScript (ES6), 36 28 27 bytes

^{Saved 1 byte by inverting the result, as suggested by LarsW}

Takes input in lowercase. Returns false for a strong word and true for a non-strong word.

s=>/[^aeiouy]{2}/.test(s+0)

How?

We append a 0 (non-vowel) at the end of the input string and look for two consecutive non-vowel characters. This allows us to cover both cases that make a word not strong:

• it contains two consecutive consonants
• or it ends with a consonant

Test cases

let f =

s=>/[^aeiouy]{2}/.test(s+0)

;[
  "hate", "love", "popularize", "academy", "you", "mouse", "a", "euouae",
  "acorn", "nut", "ah", "strong", "false", "parakeet"
]
.forEach(s => console.log(s + ' --> ' + f(s)))

Answer 2

Python 2, 48 bytes

lambda s:'se, F'in`[v in'aeiouy'for v in s+'b']`

An unnamed function taking a (lowercase) string, s, and returning False if strong or True if not.

Try it online! (inverts the results to match the OP)

How?

Non-strong words have either a consonant followed by a consonant or end in a consonant.

The code adds a consonant to the end (s+'b') to make the required test be just for two consonants in a row.

It finds out if each letter in the altered word is a vowel with the list comprehension [v in'aeiouy'for v in s+'b'].

It now needs to check for two False results in a row (signalling a non-strong word), it does so by getting a string representation (using `...`) of this list and looking for the existence of 'se, F'. This is the shortest string found in 'False, False' but none of: 'True, True'; 'False, True'; or 'True, False'.

As an example consider 'nut', the list comprehension evaluates each letter, v, of 'nutb' for existence in 'aeiouy' yielding the list [False, True, False, False], the string representation of this list is '[False, True, False, False]' which contains 'e, F' here: '[False, True, Fals>>e, F<<alse]' hence the function returns True meaning that nut is not a strong word.

Answer 3

Jelly,  10  9 bytes

e€ØY;Ạ11ẇ

A monadic link taking a list of characters and returning:

• 0 if strong
• 1 if not

Try it online! or see the test-suite.

How?

e€ØY;Ạ11ẇ - Link: list of characters, s      e.g. "hate"  or  "you"  or  "not"
  ØY      - consonant yield                   "BCDFGHJKLMNPQRSTVWXZbcdfghjklmnpqrstvwxz"
e€        - exists in? for €ach letter            [1,0,1,0]   [0,0,0]    [1,0,1]
     Ạ    - all truthy? (1 for any valid input)   1           1          1
    ;     - concatenate                           [1,0,1,0,1] [0,0,0,1]  [1,0,1,1]
      11  - literal eleven
        ẇ - sublist exists?                       0           0          1
          -  N.B.: left of ẇ implicitly makes digits so it looks for the sublist [1,1]

Note: The reason for using Ạ is just to save a byte over using 1 (since we then want to use 11 straight away).

Answer 4

R, 43 bytes

function(s)grep("[^aeiouy]{2}",paste(s,""))

Try it online!

A port of Arnauld's JavaScript answer; returns 1 for weak words and integer(0) for strong ones; it appends a (space) to the end of the string.

This is actually vectorized; with a vector of strings, it returns the indices (1-based) of the weak words.

Answer 5

05AB1E, 8 bytes

Code

žPS¡¦õÊP

Uses the 05AB1E encoding. Try it online!

Explanation

žPS¡         # Split the string on consonants (bcdfghjklmnpqrstvwxz)
    ¦        # Remove the first element of the array to handle cases when the
               string starts with a consonant
     õÊP     # Check if the empty string is not in the array

Example

             # "popularize"
žPS¡         # ['', 'o', 'u', 'a', 'i', 'e']
    ¦        # ['o', 'u', 'a', 'i', 'e']
     õÊ      # [1, 1, 1, 1, 1]
       P     # 1

Answer 6

Dyalog APL, 20 bytes

⎕←∧/2∨/0,⍨⍞∊'aeiouy'

Try it online!

Answer 7

Haskell, 61 54 bytes

f=and.(zipWith(||)=<<tail).(map(`elem`"aeiouy")).(++"z")

Try it online!

I had to add a z at the end of the string to handle the case of a trailing consonant.

Answer 8

Java (OpenJDK 8), 93 81 bytes

s->{int w=0,p=w,l;for(char c:s)w|=p&(p=l="aeiouy".indexOf(c)>>31);return w+p>=0;}

Try it online!

Answer 9

Pyth, 18 bytes

:+Q1."2}M>åYà

Verify all the test cases.

"Borrowed" the regex from the JS answer. This returns False for strong words, True otherwise

Answer 10

Husk, 12 bytes

ΛΣX_2m€¨γaıu

Try it online!

Thanks to H.PWiz for help with -4. Returns inconsistent but appropriately truthy or falsy values.
Thanks to Leo for -1, now returns consistent truthy/falsy value.

Answer 11

Brachylog, 18 11 10 bytes

,Ḷs₂{¬∈Ẉ}ᵐ

Try it online!

Neat and simple (except maybe for the 2 extra initial bytes to handle the final consonant case, like "parakeet").

Is falsey for strong words and truthy for non-strong words.

,Ḷ               % append a newline (non-vowel) at the end of input, 
                 %   to catch final consonants
     s₂          % the result has some substring of length 2
       {¬∈Ẉ}ᵐ    % where neither of its elements belong to  
                 %   the set of alternate vowels (with "y")

Answer 12

Python 2, 58 bytes

-30 bytes by realizing it can be as simple as Arnauld's JS answer.

lambda s:re.search('[^aeiouy]([^aeiouy]|$)',s)<1
import re

Try it online!

Answer 13

Jelly, 11 bytes

e€ØY;1a2\¬Ȧ

Try it online!

e€ØY;1a2\¬Ȧ  Main link
 €           For each letter
e            Is it an element of
  ØY         The consonants (excluding Yy)?
    ;1       Append 1 (true) (consonant) to make sure last letter isn't consonant
       2\    For all (overlapping) slices of length 2 (the <link><nilad>\ functionality)
      a      Logical AND of the two values; is it a consonant pair?
         ¬   Logical NOT vectorizing; for each (overlapping) pair, is it not a consonant pair?
          Ȧ  Any and all; make sure all pairs are not consonant pairs

Yes I know I've been beaten a lot by Jonathan Allan but I wanted to share my approach anyway :P

-4 bytes by stealing a little bit of Jonathan Allan's answer (instead of appending a consonant to check for last-letter edge case, just append 1)
-1 byte thanks to miles

Answer 14

Perl 5, 31 bytes (30 + 1)

$_=''if/[^aeiouy](?![aeiouy])/

+1 byte for -p command line flag. Prints the word if it's a strong word, or the empty string if it is not.

Answer 15

Retina, 23 18 bytes

$
$
1`[^aeiouy]{2}

Try it online! Outputs 0 for strong, 1 if not. Add 1 byte to support mixed case. Edit: Saved 5 bytes thanks to @ovs.

Answer 16

Awk, 39 bytes

/([^aeiouy]{2}|[^aeiouy]$)/{print "n"}

prints n for non-strongword, nothing (or, just a newline) for strongword

following the pack and searching for two consecutive non-vowels on lowercase input

testing

$ awk -f strongwork.awk
hate
love
popularize
academy
you
mouse
acorn
n
nut
n
ah
n
strong
n
false
n
parakeet
n

Answer 17

Kotlin, 49 bytes

{Regex(".*[^aeiouy]([^aeiouy].*|$)").matches(it)}

True and false are swapped

Beautified

{
    Regex(".*[^aeiouy]([^aeiouy].*|$)").matches(it)
}

Test

var s:(String)->Boolean =
{Regex(".*[^aeiouy]([^aeiouy].*|$)").matches(it)}
data class TestData(val input: String, val output: Boolean)

fun main(args: Array<String>) {
    val items = listOf(
            TestData("hate", true),
            TestData("love", true),
            TestData("popularize", true),
            TestData("academy", true),
            TestData("you", true),
            TestData("mouse", true),
            TestData("acorn", false),
            TestData("nut", false),
            TestData("ah", false),
            TestData("strong", false),
            TestData("false", false),
            TestData("parakeet", false)
    )

    items
            .filter { s(it.input) == it.output }
            .forEach { throw AssertionError(it.toString()) }

    println("Test Passed")
}

TryItOnline

Based on @Arnauld's Answer

Answer 18

Java 8, 53 42 bytes

s->s.matches(".*[^aeiouy]([^aeiouy].*|$)")

-11 bytes by using the same regex as in @jrtapsell's Kotlin answer instead.

Try it here. (false if strong; true if not)

Explanation:

s->               // Method with String parameter and boolean return-type
  s.matches(      //  Checks if the String matches the following regex:
    ".*           //   One or more characters
     [^aeiouy]    //   Followed by a consonant
     ([^aeiouy].* //   Followed by another consonant (+ any more characters)
      |$)")       //   Or the end of the String
                  // End of method (implicit / single-line return statement)

So it basically checks if we can find two adjacent consonants, or if the String ends with a consonant.

---

Old answer (53 bytes):

s->s.matches("[aeiouy]*([a-z&&[^aeiouy]][aeiouy]+)*")

Try it here. (true if strong; false if not)

Uses regex to see if the input-String matches the 'strong'-regex. Note that String#matches in Java automatically adds ^...$ to check if the String entirely matches the given regex.

Explanation":

 s->                   // Method with String parameter and boolean return-type
  s.matches(           //  Checks if the String matches the following regex:
    "[aeiouy]*         //   0 or more vowels
    ([a-z&&[^aeiouy]]  //     { A consonant,
     [aeiouy]+)        //       plus one or more vowels }
    *")                //    Repeated 0 or more times
                       // End of method (implicit / single-line return statement)

---

A search instead of matches (like a lot of other answers use) is actually longer in Java:
70 bytes:

s->java.util.regex.Pattern.compile("[^aeiouy]{2}").matcher(s+0).find()

Try it here. (false if strong; true if not)

Answer 19

Ruby, 25 bytes

->w{w+?t!~/[^aeiouy]{2}/}

Try it online!

Everybody else is doing it, so why can't Ruby?

Answer 20

SOGL V0.12, 19 18 bytes

æ"[^ŗy]”ŗ(ŗ|$)”øβ=

Try it Here!

Explanation:

æ                   push "aeiou"
 "[^ŗy]”            push "[^ŗy]" with ŗ replaced with pop
        ŗ(ŗ|$)”     push `ŗ(ŗ|$)` with ŗ replaced with pop
               øβ   replace in the input that regex with nothing
                 =  check for equality with the original input

Answer 21

05AB1E, 11 bytes

žP¹SåJ1«11å

Try it online!

Uses Jonathan's algorithm, returns 0 for true and 1 for false.

Answer 22

C++, 195 194 bytes

-1 bytes thanks to Zacharý

Uppercase, return true if input is a strong word, false otherwise ( C++ have simple int to bool implicit cast rules, 0 => false, true otherwise )

#include<string>
#define C(p)(v.find(e[p])==size_t(-1))
std::string v="AEIOUY";int s(std::string e){for(int i=0;i<e.size()-1;++i)if(e[i]>64&&e[i]<91&&C(i)&&C(i+1))return 0;return!C(e.size()-1);}

Code to test :

auto t = {
    "HATE",
    "LOVE",
    "POPULARIZE",
    "ACADEMY",
    "YOU",
    "MOUSE",
    "ACORN",
    "NUT",
    "AH",
    "STRONG",
    "FALSE",
    "PARAKEET"
};

for (auto&a : t) {
    std::cout << (s(a) ? "true" : "false") << '\n';
}

Answer 23

Swift 3.1, 85 bytes

import Foundation
{($0+"0").range(of:"[^aeiouy]{2}",options:.regularExpression)==nil}

Try it here!

This borrows Arnauld's regex.

Answer 24

Lua, 41 bytes

return#(io.read()..0):match"[^aeiouy]+"<2

Reads from standard input

Lua (loadstring'ed), 37 bytes

return#((...)..0):match"[^aeiouy]+"<2

Reads from function parameter(s)

---

Input is lowercase

Sees if there is a string of length 2 or more, consisting only of not vowels (consonants) or if the string ends with a non-vowel

Returns true/false

Answer 25

C, 107 Bytes

i,v,w,r,t;a(char*s){w=0;for(r=1;*s;s++){v=1;for(i=6;v&&i;)v=*s^" aeiouy"[i--];r=w&&v?0:r;w=v;}return r&~v;}

Returns 1 for strong word and 0 for weak word. Tested with the words given in the main post.

Answer 26

C (gcc), 59 bytes

f(char*s){return*s?strcspn(s,"aeiouy")+!s[1]<2?f(s+1):0:1;}

Try it online!

Answer 27

PHP, 69 bytes

preg_match("/([^AEIOUY][^AEIOUY]+|[^AEIOUY]$)/",$_SERVER['argv'][1]);

Returns 1 is the word is not strong.

Answer 28

CJam, 57 bytes

q{"aeiouy"#W=}%_,:B{_A={_A_)\B(<{=!X&:X;}{0:X;;;}?}&}fA;X

Try it online!

---

Reads input, converts to 1s for consonants, 0s for vowels. For every consonant, AND predefined variable X (predefined to 1) with next character's value. Output X