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 <aniket...@gmail.com> 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 algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com. > 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 algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
