Given a dictionary and a list of letters find all valid words that can be built with the letters

Question

The brute force way can solve the problem in O(n!), basically calculating all the permutations and checking the results in a dictionary. I am looking for ways to improve the com

Answer 1

I was recently asked the same question in BankBazaar interview. I was given the option to (he said that in a very subtle manner) pre-process the dictionary in any way I want.

My first thought was to arrange the dictionary in a trie or ternary search tree, and make all the words from the letters given. In any optimization way, that would take n! + n-1! + n-2! n-3! + ..... + n word checks(n being the number of letters) in worst case, which was not acceptable.

The other way could be to check all the dictionary words whether they can be made from the given letters. This again in any optimized way would take noOfDictionaryWords(m) * average size of dictionary words(k) at worst case, which was again not acceptable.

Now I have n! + n-1! + n-2! + .... + N words, which I have to check in the dictionary, and I don't want to check them all, so what are the situations that I have to check only a subset of them, and how to group them.

If I have to check only combination and not permutation, the result gets to 2^n.

so I have to pre-process the dictionary words in such a way that if I pass a combination, all the anagrams would be printed.

A ds something like this : http://1.bp.blogspot.com/-9Usl9unQJpY/Vg6IIO3gpsI/AAAAAAAAAbM/oTuhRDWelhQ/s1600/hashmapArrayForthElement.png A hashvalue made by the letters(irrespective of its positions and permutation), pointing to list containing all the words made by those letters, then we only need to check that hashvalue.

I gave the answer to make the hash value by assigning a prime value to all the alphabets and while calculating the hash value of a word, multiply all the assigned values. This will create a problem of having really big hash values given that 26th prime is 101, and many null values in the map taking space. We could optimize it a bit by rather than starting lexicographically with a = 2, b = 3, c = 5, d = 7.... z = 101, we search for the most used alphabets and assign them small values, like vowels, and 's', 't' etc. The interviewer accepted it, but was not expecting the answer, so there is definitely another answer, for better or worse but there is.