Merge pull request #116 from jsca-kwok/jk-anagram

Anagram: sorting approach

DIRECTORY.md

@@ -18,6 +18,7 @@
     * [Single Number](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/single_number.rb)
     * [Sort Squares Of An Array](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/sort_squares_of_an_array.rb)
     * Strings
+      * [Anagram Checker](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/strings/anagram_checker.rb)
       * [Jewels And Stones](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/strings/jewels_and_stones.rb)
       * [Remove Vowels](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/strings/remove_vowels.rb)
     * [Two Sum](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/two_sum.rb)

data_structures/arrays/strings/anagram_checker.rb

@@ -0,0 +1,45 @@
+# Challenge name: Is anagram
+#
+# Given two strings s and t , write a function to determine
+# if t is an anagram of s.
+#
+# Note:
+# You may assume the string contains only lowercase alphabets.
+#
+# Follow up:
+# What if the inputs contain unicode characters?
+# How would you adapt your solution to such case?
+#
+# @param {String} s
+# @param {String} t
+# @return {Boolean}
+
+#
+# Approach 1: Sort and Compare
+#
+# Complexity analysis:
+#
+# Time Complexity: O(n log n). Assume that n is the length of s, sorting costs O(n log n), and comparing two strings costs O(n). Sorting time dominates and the overall time complexity is O(n log n).
+# Space Complexity: O(1). Space depends on the sorting implementation which, usually, costs O(1) auxiliary space if heapsort is used.
+#
+def is_anagram(s, t)
+  return false if s.length != t.length
+
+  arr1 = s.split('').sort
+  arr2 = t.split('').sort
+
+  arr1 == arr2
+end
+
+s = 'anagram'
+t = 'nagaram'
+puts(is_anagram(s, t))
+# => true
+s = 'rat'
+t = 'car'
+puts(is_anagram(s, t))
+# => false
+s = 'a'
+t = 'ab'
+puts(is_anagram(s, t))
+# => false