How to resolve the algorithm Anagrams/Deranged anagrams step by step in the Erlang programming language

Published on 12 May 2024 09:40 PM
#Erlang

How to resolve the algorithm Anagrams/Deranged anagrams step by step in the Erlang programming language

Table of Contents

Problem Statement

Two or more words are said to be anagrams if they have the same characters, but in a different order. By analogy with derangements we define a deranged anagram as two words with the same characters, but in which the same character does not appear in the same position in both words. Use the word list at unixdict to find and display the longest deranged anagram.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Anagrams/Deranged anagrams step by step in the Erlang programming language

Source code in the erlang programming language

-module( anagrams_deranged ).
-export( [task/0, words_from_url/1] ).

task() ->
       find_unimplemented_tasks:init_http(),
       Words = words_from_url( "http://www.puzzlers.org/pub/wordlists/unixdict.txt" ),
       Anagram_dict = anagrams:fetch( Words, dict:new() ),
       Deranged_anagrams = deranged_anagrams( Anagram_dict ),
       {_Length, Longest_anagrams} = dict:fold( fun keep_longest/3, {0, []}, Deranged_anagrams ),
       Longest_anagrams.

words_from_url( URL ) ->
	{ok, {{_HTTP, 200, "OK"}, _Headers, Body}} = httpc:request( URL ),
	string:tokens( Body, "\n" ).


deranged_anagrams( Dict ) ->
        Deranged_dict = dict:map( fun deranged_words/2, Dict ),
        dict:filter( fun is_anagram/2, Deranged_dict ).

deranged_words( _Key, [H | T] ) ->
        [{H, X} || X <- T, is_deranged_word(H, X)].

keep_longest( _Key, [{One, _} | _]=New, {Length, Acc} ) ->
        keep_longest_new( erlang:length(One), Length, New, Acc ).

keep_longest_new( New_length, Acc_length, New, _Acc ) when New_length > Acc_length ->
        {New_length, New};
keep_longest_new( New_length, Acc_length, New, Acc ) when New_length =:= Acc_length ->
        {Acc_length, Acc ++ New};
keep_longest_new( _New_length, Acc_length, _New, Acc ) ->
        {Acc_length, Acc}.

is_anagram( _Key, [] ) -> false;
is_anagram( _Key, _Value ) -> true.

is_deranged_word( Word1, Word2 ) ->
        lists:all( fun is_deranged_char/1, lists:zip(Word1, Word2) ).

is_deranged_char( {One, Two} ) -> One =/= Two.

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