yeah, that is normal bryteforce. Any better idea? On 11/14/11, Ankur Garg <[email protected]> wrote: > We can use a trie here .. Create a trie with all words of dictionary . > > Now delete the last character of the word and check if such a word is a > valid word . If not see if adding a new character can make it a valid word > . If not delete the next character and repeat the process again . > > This is what I can think of here. Any other solutions/guesses ? > > > > On Mon, Nov 14, 2011 at 12:43 PM, Aniket <[email protected]> wrote: > >> You are given a word and a dictionary. Now propose an algorithm edit >> the word (insert / delete characters) minimally to get a word that >> also exists in the dictionary. Cost of insertion and deletion is same. >> Write pseudocode for it. >> >> Seems like minimum edit distance problem but some modification is >> needed. >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to [email protected]. >> To unsubscribe from this group, send email to >> [email protected]. >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> >> > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > >
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