Question

Two words are anagrams if one of them has exactly same characters as that of the another word.

Example : Anagram & Nagaram are anagrams (case-insensitive).

Now there are many questions similar to this . A couple of approaches to find whether two strings are anagrams are :

1) Sort the strings and compare them.

2) Create a frequency map for these strings and check if they are the same or not.

But in this case , we are given with a word (for the sake of simplicity let us assume a single word only and it will have single word anagrams only) and we need to find anagrams for that.

Solution which I have in mind is that , we can generate all permutations for the word and check which of these words exist in the dictionary . But clearly , this is highly inefficient. Yes , the dictionary is available too.

So what alternatives do we have here ?

I also read in a similar thread that something can be done using Tries but the person didn’t explain as to what the algorithm was and why did we use a Trie in first place , just an implementation was provided that too in Python or Ruby. So that wasn’t really helpful which is why I have created this new thread. If someone wants to share their implementation (other than C,C++ or Java) then kindly explain it too.

Answer 1

Example algorithm:

Open dictionary
Create empty hashmap H
For each word in dictionary:
  Create a key that is the word's letters sorted alphabetically (and forced to one case)
  Add the word to the list of words accessed by the hash key in H

To check for all anagrams of a given word:

Create a key that is the letters of the word, sorted (and forced to one case)
Look up that key in H
You now have a list of all anagrams

Relatively fast to build, blazingly fast on look-up.