From 4e023f7a0b55ff78c5528de7dedc848156566820 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Jobsamuel=20N=C3=BA=C3=B1ez?= Date: Tue, 12 Dec 2017 11:54:50 -0400 Subject: [PATCH] refactor(anagrams): improve code legibility --- README.md | 11 +++++------ snippets/anagrams-of-string-(with-duplicates).md | 10 +++++----- 2 files changed, 10 insertions(+), 11 deletions(-) diff --git a/README.md b/README.md index dee253d4a..f352976b5 100644 --- a/README.md +++ b/README.md @@ -59,11 +59,11 @@ Use `map()` to combine the letter with each partial anagram, then `reduce()` to Base cases are for string `length` equal to `2` or `1`. ```js -const anagrams = s => { - if(s.length <= 2) return s.length === 2 ? [s, s[1] + s[0]] : [s]; - return s.split('').reduce( (a,l,i) => { - anagrams(s.slice(0,i) + s.slice(i+1)).map( v => a.push(l+v) ); - return a; +const anagrams = str => { + if(str.length <= 2) return str.length === 2 ? [str, str[1] + str[0]] : [str]; + return str.split('').reduce( (acc, letter, index) => { + anagrams(str.slice(0, index) + str.slice(index + 1)).map( value => acc.push(letter + value) ); + return acc; }, []); } ``` @@ -459,4 +459,3 @@ const validateNumber = n => !isNaN(parseFloat(n)) && isFinite(n); ## Credits *Icons made by [Smashicons](https://www.flaticon.com/authors/smashicons) from [www.flaticon.com](https://www.flaticon.com/) is licensed by [CC 3.0 BY](http://creativecommons.org/licenses/by/3.0/).* - diff --git a/snippets/anagrams-of-string-(with-duplicates).md b/snippets/anagrams-of-string-(with-duplicates).md index 97140f0e4..16c1a80f6 100644 --- a/snippets/anagrams-of-string-(with-duplicates).md +++ b/snippets/anagrams-of-string-(with-duplicates).md @@ -6,11 +6,11 @@ Use `map()` to combine the letter with each partial anagram, then `reduce()` to Base cases are for string `length` equal to `2` or `1`. ```js -const anagrams = s => { - if(s.length <= 2) return s.length === 2 ? [s, s[1] + s[0]] : [s]; - return s.split('').reduce( (a,l,i) => { - anagrams(s.slice(0,i) + s.slice(i+1)).map( v => a.push(l+v) ); - return a; +const anagrams = str => { + if(str.length <= 2) return str.length === 2 ? [str, str[1] + str[0]] : [str]; + return str.split('').reduce( (acc, letter, index) => { + anagrams(str.slice(0, index) + str.slice(index + 1)).map( value => acc.push(letter + value) ); + return acc; }, []); } ```