Generate same unique hash code for all anagrams

Hash function : Assign primary numbers to each character. While calculating hash code, get the prime number assigned to that character and multiply with to existing value.Now all anagrams produce same hash value.

ex : a - 2, c - 3 t - 7

hash value of cat = 3*2*7 = 42 hash value of act = 2*3*7 = 42 Print all strings which are having same hash value(anagrams will have same hash value)

Hashing with the sorted string is pretty nice, i'd have done that probably, but it could indeed be slow and cumbersome. Here's another thought, not sure if it works - pick a set of prime numbers, as small as you like, the same size as your character set, and build a fast mapping function from your chars to that. Then for a given word, map each character into the matching prime, and multiply. finally, hash using the result.

This is very similar to what Heuster suggested, only with less collisions (actually, I believe there will be no false collisions, given the uniqueness of the prime decomposition of any number).

simple e.g. -

int primes[] = {2, 3, 5, 7, ...} // can be auto generated with a simple code

inline int prime_map(char c) {
    // check c is in legal char set bounds
    return primes[c - first_char];
}

...
char* word = get_next_word();
char* ptr = word;
int key = 1;
while (*ptr != NULL) {
    key *= prime_map(*ptr);
    ptr++;
}
hash[key].add_to_list(word); 

[edit]

A few words about the uniqueness - any integer number has a single breakdown to multiplications of primes, so given an integer key in the hash you can actually reconstruct all possible strings that would hash to it, and only these words. Just break into primes, p1^n1*p2^n2*... and convert each prime to the matching char. the char for p1 would appear n1 times, and so on. You can't get any new prime you didn't explicitly used, being prime means you can't get it by any multiplication of other primes.

This brings another possible improvement - if you can construct the string, you just need to mark the permutations you saw when populating the hash. since the permutations can be ordered by lexicographic order, you can replace each one with a number. This would save the space of storing the actual strings in the hash, but would require more computations so it's not necessarily a good design choice. Still, it makes a nice complication of the original question for interviews :)

The other posters suggested converting characters into prime numbers and multiplying them together. If you do this modulo a large prime, you get a good hash function that won't overflow. I tested the following Ruby code against the Unix word list of most English words and found no hash collisions between words that are not anagrams of one another. (On MAC OS X, this file is located here: /usr/share/dict/words.)

My word_hash function takes the ordinal value of each character mod 32. This will make sure that uppercase and lowercase letters have the same code. The large prime I use is 2^58 - 27. Any large prime will do so long as it is less than 2^64 / A where A is my alphabet size. I am using 32 as my alphabet size, so this means I can't use a number larger than about 2^59 - 1. Since ruby uses one bit for sign and a second bit to indicate if the value is a number or an object, I lose a bit over other languages.

def word_hash(w)
  # 32 prime numbers so we can use x.ord % 32. Doing this, 'A' and 'a' get the same hash value, 'B' matches 'b', etc for all the upper and lower cased characters.
  # Punctuation gets assigned values that overlap the letters, but we don't care about that much.
  primes = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131]
  # Use a large prime number as modulus. It must be small enough so that it will not overflow if multiplied by 32 (2^5). 2^64 / 2^5 equals 2^59, so we go a little lower.
  prime_modulus = (1 << 58) - 27
  h = w.chars.reduce(1) { |memo,letter| memo * primes[letter.ord % 32] % prime_modulus; }
end

words = (IO.readlines "/usr/share/dict/words").map{|word| word.downcase.chomp}.uniq
wordcount = words.size
anagramcount = words.map { |w| w.chars.sort.join }.uniq.count

whash = {}
inverse_hash = {}
words.each do |w|
  h = word_hash(w)
  whash[w] = h
  x = inverse_hash[h]
  if x && x.each_char.sort.join != w.each_char.sort.join
    puts "Collision between #{w} and #{x}"
  else
    inverse_hash[h] = w
  end
end
hashcount = whash.values.uniq.size
puts "Unique words (ignoring capitalization) = #{wordcount}. Unique anagrams = #{anagramcount}. Unique hash values = #{hashcount}."