finding if two words are anagrams of each other

Count the frequency of each character in the two strings. Check if the two histograms match. O(n) time, O(1) space (assuming ASCII) (Of course it is still O(1) space for Unicode but the table will become very large).

Get table of prime numbers, enough to map each prime to every character. So start from 1, going through line, multiply the number by the prime representing current character. Number you'll get is only depend on characters in string but not on their order, and every unique set of characters correspond to unique number, as any number may be factored in only one way. So you can just compare two numbers to say if a strings are anagrams of each other.

Unfortunately you have to use multiple precision (arbitrary-precision) integer arithmetic to do this, or you will get overflow or rounding exceptions when using this method.
For this you may use libraries like BigInteger, GMP, MPIR or IntX.

Pseudocode:

prime[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101}

primehash(string)
    Y = 1;
    foreach character in string
        Y = Y * prime[character-'a']

    return Y

isanagram(str1, str2)
    return primehash(str1)==primehash(str2)

1. Create a Hashmap where key - letter and value - frequencey of letter,
2. for first string populate the hashmap (O(n))
3. for second string decrement count and remove element from hashmap O(n)
4. if hashmap is empty, the string is anagram otherwise not.