yes levenshtein distance and BK tree can be used to solve this. where edge weight between nodes is equal to levenshtein distance.
On Wed, Nov 23, 2011 at 7:14 PM, abhishek kumar <afs.abhis...@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. > > > -- > Abhishek Kumar > B.Tech(IT) Graduate > Allahabad > Contact no-+919663082731 > > -- > 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.
