Is it a near-anagram?
Two words are anagrams of each other if the letters of one can be reordered to spell the other; e.g. ADOBE and ABODE are anagrams. An alternate way of describing it is that both words contain the same count of each letter. If you were to make a table:
ADOBE ABODE
----- -----
A: 1 A: 1
B: 1 B: 1
D: 1 D: 1
E: 1 E: 1
O: 1 O: 1
We can define a "near-anagram" as a pair of words that are almost anagrams, in the sense that they differ by only one letter. For example, TULIP and TUPLE are near-anagrams. TUPLE can be rearranged to spell TULEP, which differs from TULIP by only one letter. As a table:
TUPLE TULIP
----- -----
E: 1 I: 1
L: 1 L: 1
P: 1 P: 1
T: 1 T: 1
U: 1 U: 1
Challenge
The challenge is to take two strings as input and determine if they are near-anagrams.
- The strings can be taken in any convenient format for your language (strings, sequences of characters, etc.)
- The output can be any two distinct values, as long as they are always consistent; e.g. 0 and 1 for false and true.
- The strings will only contain alphabet characters in a single case. You can assume either upper or lower, whichever is convenient; examples are in upper. Input will contain no whitespace. (It is acceptable to take the input as a single string containing the words separated by whitespace, if it is convenient.)
- You can assume the strings will not be the same. A decision problem to handle equal strings is not very interesting. They may be proper anagrams, however.
- You can assume the input is not empty.
- The two words might not be the same length; they may differ by one letter at most (see test cases.)
Winning criteria is code-golf. Shortest answer in each language wins.
Test Cases
ADOBE ABODE -> false (proper anagram)
TUPLE TULIP -> true (near-anagram)
ABCDE DADBC -> true (two Ds)
BAR BARN -> true (one extra letter)
BARN BARREN -> false (too different)
[APL (Dyalog Extended)], 18 by …
3y ago
[JavaScript (Node.js)], 131 by …
3y ago
Stax, 12 bytes ä╫◙=♥:≡ƒélΣ …
3y ago
[Brachylog], 8 bytes pᵐ …
3y ago
[JavaScript (Node.js)], 150 71 …
3y ago
5 answers
You are accessing this answer with a direct link, so it's being shown above all other answers regardless of its score. You can return to the normal view.
APL (Dyalog Extended), 18 bytes
Anonymous tacit prefix function taking a list of two strings as argument.
2=1⊥(≠⌿∊≢⍤⊢⌸⍤,⍤1↑)
(…) apply the following tacit function to the argument:
↑ mix the list of strings into a 2-row character matrix (padding any short string with spaces as necessary)
∊…⍤1 with the enlisted (flattened into a single string) argument as left argument, apply the following function to each row of the matrix:
⍤, concatenate the enlisted string to the row string, then:
≢⍤⊢⌸ count the number of occurrences of each unique element, in their order of appearance
≠⌿ Boolean mask indicating where the rows differ
1⊥ sum (lit. evaluate as unary; i.e. count differences)
2= is 2 equal to that?
0 comment threads
JavaScript (Node.js), 131 bytes
Returns true and false
g=(s,t,c=[...t])=>([...s].forEach(v=>(y=c.indexOf(v),c.splice(y,y>=0))),c)
f=(s,t)=>(x=g(s,t).length)<2&&(y=g(t,s).length)<2&&x+y
Explanation
g computes a sort of set (Array?) difference between t and s, running through the characters of s and removing any matches in t, then returning the remaining characters in t.
f calls g to compute both s-t and t-s and checks that both are less than 2 (i.e. have only one letter difference either way). Then it checks the sum of the difference is non-zero to check that there is a difference, which I needed to add in to make perfect anagrams false.[1]
-
Without the restriction on perfect anagrams being falsey, I could have removed all of the x and y logic to get a 116 char solution ↩︎
0 comment threads
Brachylog, 8 bytes
pᵐ≠{b|}ᵛ
Since it takes $5!*5! = 14400$ tries to determine that the first test case gives false, I've added some shorter ones...
pᵐ Permute each element of the input.
≠ The permuted elements are not equal,
{ }ᵛ but yield the same result if
| maybe
b their first elements are removed.
0 comment threads
JavaScript (Node.js), 150 71 bytes
f=(a,b)=>a[b.length]?f(b,a):b.filter(x=>a[y=a.indexOf(x)]=!~y).length-1
1 comment thread