caN it be acrOnymised?

Question

The Typical Way to Make an Acronym Out of a Phrase Is to Take the First Letter of Each Word: TTWMAOPITFLEW. howeveR, sometimEs, you can make an acronym of random leTters In a seNtence such As like this: RETINA. The only condition is that the letters have to be in the correct order. For instance:

• LORD can be acronymised from Hello World: heLlO woRlD
• LEOD cannot be acronymised from Hello World, as no ls are before e

Your task is to take two strings as input as to determine if one can be acronymised into the other.

The first input, the phrase, will only contain letters (A-Z or a-z) and spaces, and the second input, the acronym, will only contain letters (A-Z or a-z). The acronym will always be shorter, or of equal size, to the phrase, and both the acronym and the phrase will be, at minimum, 1 letter long. You may choose which case (upper or lower) you want the inputs to be.

You may choose any two values to represent true and false, as long as those values are consistent.

This is code-golf so the shortest code in bytes wins

Examples

HELLO WORLD, LORD -> true
CODE GOLF AND CODING CHALLENGES, DANGLE -> true
SANDBOX FOR PROPOSED CHALLENGES, CODE -> false
HELLO WORLD, LLLD -> true
HELLO WORLD, LLDL -> false
NEW YORK POLICE DEPARTMENT, NOODLE -> false
MASSACHUSETTS INSTITUTE OF TECHNOLOGY, MUTTON -> true
BOB, BOB -> true
PRESIDENT OF THE UNITED STATES, I -> true

Answer 1

Brachylog, 1 byte

⊇

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Answer 2

Java 8, 46 39 38 bytes

a->b->a.matches(b.replaceAll("",".*"))

-7 bytes thanks to @tsh.
-1 byte thanks to @NahuelFouilleul.

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Explanation:

a->b->         // Method with two String parameters and boolean return-type
  a.matches(   //  Check if the first input matches the regex:
   b           //   The second input,
    .replaceAll("",".*"))
               //   where every character is surrounded with ".*"

For example:

a="HELLO WORLD"
b="LORD"

Will do the check:

"HELLO WORLD".matches("^.*L.*O.*R.*D.*$")

(The ^...$ will add the String#matches builtin implicitly, since it will always try to match the entire String.)

Answer 3

C (gcc), 47 bytes

f(a,b)char*a,*b;{a=!*b||*a&&f(a+1,b+(*a==*b));}

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Answer 4

05AB1E, 3 bytes

æIå

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æ              # power set of the first input
 I             # second input
  å            # does a contain b?
               # implicit output

Answer 5

Python 2, 48 bytes

lambda s,a:re.search('.*'.join(a),s)>0
import re

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Answer 6

APL (Dyalog Unicode), 17 bytes

Full program. Prompts for phrase, then acronym.

0∊⊃(⍳⍨↓⊢)/⍞,⊂⌽0,⍞

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⍞ prompt for phrase
  "HELLO WORLD"
  "HELLO WORLD"

0, prepend a zero
  [0,'H','E','L','L','O',' ','W','O','R','L','D']
  [0,'H','E','L','L','O',' ','W','O','R','L','D']

⌽ reverse
  ['D','L','R','O','W',' ','O','L','L','E','H',0]
  ['D','L','R','O','W',' ','O','L','L','E','H',0]

⊂ enclose (to treat as a whole)
  [['D','L','R','O','W',' ','O','L','L','E','H',0]]
  [['D','L','R','O','W',' ','O','L','L','E','H',0]]

⍞, prepend the prompted-for acronym:
  ['L','L','L','D',['D','L','R','O','W',' ','O','L','L','E','H',0]]
  ['L','L','D','L',['D','L','R','O','W',' ','O','L','L','E','H',0]]

(…)/ reduce that list by the following tacit function:

  ⍳⍨ the ɩndex of the first occurrence in the phrase (will return 1 + phrase length if unfound)

  ↓⊢ drop that many characters from the phrase

We then use the shortened phrase to look for the next letter. If at any point a letter is unfound, we'll drop everything, including the final zero. This means that if our acronym is good, we'll still have a zero left.
  [0]
  []

⊃ disclose (because the reduction reduced the number of dimensions from 1 to 0)

0∊ is zero a member thereof?

Answer 7

J, 21 bytes

<@[e.]<@#~2#:@i.@^#@]

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Note: Some of the longer test cases omitted because they this solution is O(2^n). They would pass with infinite memory.

Explanation: We create all 2 ^ (length of haystack) possible substrings, and check if the needle is an element of that list.

Answer 8

Gaia, 5 bytes

$z@$Ė

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Powerset approach.

Answer 9

JavaScript, 26 bytes

Port of Kevin's Java solution so please +1 him, too.

Takes the string as a string via parameter s and the acronym as a character array via parameter a. Outputs false for true and true for false.

s=>a=>!s.match(a.join`.*`)

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Answer 10

Python 3, 89 68 bytes

-21 bytes thanks to AdmBorkBork

def a(b,c):
    for i in b:
        if i==c[:1]:
            c=c[1:]
    return len(c)==0

Takes input as a(string, acronym). Try it online!

Answer 11

Japt, 4 bytes

à øV

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Answer 12

Python 3, 63 bytes

f=lambda s,t:(t[:1]in{*s}and f(s[s.find(t[0]):],t[1:]))**len(t)

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Recursive function. Will check for each letter in the acronym t, whether it is found in the string s. If it is, the function is called recursively with the part of the string after the current test character t[0] as the new input string s.

When the test character is not found, the final evaluation result (which at that point is always 0) is raised to the power of the length of the remaining test string t. Since 0**0 == 1 and 0**x == 0 for any x > 0, the function returns 1 when all test characters have been found in order and 0 otherwise.

Answer 13

C#, 152 bytes

public class P{public static void Main(string[]a){int q=0;int e=a[1].Length;foreach(char c in a[0])if(q!=e&&c==a[1][q])q++;System.Console.Write(q==e);}}

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Answer 14

Stax, 4 bytes

äΦv>

Run and debug it

Answer 15

Jelly, 4 bytes

eŒP}

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Answer 16

Zsh, 35 bytes

[[ $1 = *${~${(j:*:)${(s::)2}}}* ]]

Outputs via exit code. Try it online!

[[ $1 = *${~${(j:*:)${(s::)2}}}* ]]
                    ${     2}        # second parameter
                    ${(s::) }        # split into characters
            ${(j:*:)         }       # join with *
         ${~                  }      # enable globbing
        *${~${(j:*:)${(s::)2}}}*     # *A*C*R*O*N*Y*M*
[[ $1 =                           ]] # does it match the first parameter?

Answer 17

JavaScript, 50 bytes

h=>n=>[...n].reduce((a,l)=>a+1?h.indexOf(l,a):a,0)

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take input as f(haystack)(needle)

h=>n=>                                             // inputs
      [...n]                                       // transform n from string to array of char
            .reduce((a,l)=>                    ,0) // for each letter change the value of a, (a starting at 0)
                           a+1?h.indexOf(l,a):a    // if a === -1 keep it else replace it by the position of the letter l in the haystack h starting from position of previous letter

Answer 18

SNOBOL4 (CSNOBOL4), 88 bytes

	T =INPUT
S	T LEN(1) . X REM . T	:F(M)
	M =M ARB X	:(S)
M	INPUT M	:F(END)
	OUTPUT =1
END

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Prints 1 for Acronymizable, and does nothing for not.

	T =INPUT			;* read in the Target
S	T LEN(1) . X REM . T	:F(M)	;* extract the first letter of T
					;* and when T is empty, goto M
	M =M ARB X	:(S)		;* create a PATTERN: M, ARBitrary match, X
					;* then goto S
M	INPUT M	:F(END)			;* if M doesn't match the input, end
	OUTPUT =1			;* else print 1
END

Answer 19

Emacs Lisp, 52 bytes

(lambda(a b)(string-match(mapconcat'string b".*")a))

Answer 20

Pyth, 3 bytes

}Ey

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Answer 21

Haskell, 40 bytes

r@(a:c)#(b:d)|a==b=c#d|1>0=r#d
x#y=x==""

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Answer 22

Retina 0.8.2, 20 19 bytes

+`(.)(.*¶)\1?
$2
¶$

Try it online! Takes the phrase and acronym on separate lines, but the link includes a header that formats the test suite appropriately. Explanation:

+`

Process all of the letters of the phrase.

(.)(.*¶)\1?
$2

For each letter of the phrase delete the next letter of the acronym if it is the same.

¶$

Check that all of the letters of the acronym were deleted.

Answer 23

Python 3, 85 bytes

def f(s,w):
	for c in s:
		if c==w[0]:
			w.pop(0)
			if len(w)==0:return 1
	return 0

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