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)

posted almost 4 years ago

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4y ago

snail_‭

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