Levensteins algorithm On 14 Nov 2011 18:19, "aniket chatterjee" <[email protected]> wrote: > > 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. > > > > > > -- > 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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