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prathamesh-sonpatki's solution

to Anagram in the Clojure Track

Published at Jul 13 2018 · 2 comments
Instructions
Test suite
Solution

Given a word and a list of possible anagrams, select the correct sublist.

Given "listen" and a list of candidates like "enlists" "google" "inlets" "banana" the program should return a list containing "inlets".

Source

Inspired by the Extreme Startup game https://github.com/rchatley/extreme_startup

Submitting Incomplete Solutions

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anagram_test.clj

(ns anagram-test
  (:require [clojure.test :refer [deftest is]]
            anagram))

(deftest no-matches
  (is (= []
         (anagram/anagrams-for "diaper" ["hello" "world" "zombies" "pants"]))))

(deftest detect-simple-anagram
  (is (= ["tan"] (anagram/anagrams-for "ant" ["tan" "stand" "at"]))))

(deftest does-not-confuse-different-duplicates
  (is (= [] (anagram/anagrams-for "galea" ["eagle"]))))

(deftest eliminate-anagram-subsets
  (is (= [] (anagram/anagrams-for "good" ["dog" "goody"]))))

(deftest detect-anagram
  (is (= ["inlets"]
         (let [coll ["enlists" "google" "inlets" "banana"]]
           (anagram/anagrams-for "listen" coll)))))

(deftest multiple-anagrams
  (is (= ["gallery" "regally" "largely"]
         (let [coll ["gallery" "ballerina" "regally"
                     "clergy"  "largely"   "leading"]]
           (anagram/anagrams-for "allergy" coll)))))

(deftest case-insensitive-anagrams
  (is (= ["Carthorse"]
         (let [coll ["cashregister" "Carthorse" "radishes"]]
           (anagram/anagrams-for "Orchestra" coll)))))

(deftest word-is-not-own-anagram
  (is (= [] (anagram/anagrams-for "banana" ["banana"]))))

(deftest capital-word-is-not-own-anagram
  (is (= [] (anagram/anagrams-for "BANANA" ["banana"]))))
(ns anagram)

(defn anagrams-for
  [word list]
  (filter (fn [w] (= (frequencies w) (frequencies word))) list))

Community comments

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Avatar of prathamesh-sonpatki
prathamesh-sonpatki
Solution Author
commented about 7 years ago

@zoldar Sorry for so late. Please check now

Avatar of zoldar

Looks nice, however you could reduce the overhead of processing word with every candidate for anagram. Also, please keep in mind that if word is exactly the same, it's not technically an anagram (test suite probably lacks this particular test).

Coming back to my original advice, I actually cancel my remark about frequencies being more efficient than sort in that particular use. A small benchmark I did some time ago proved actually slightly otherwise: https://gist.github.com/zoldar/6425609

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