this is well known problem, use the BFS traversal / Backtracking

On Nov 26, 3:54 pm, tech coder <techcoderonw...@gmail.com> wrote:
> i think edit distance algorithm can not be used here because in edit
> distance problem we have a target string and a source string. Here we dont
> have any target word.
> I think trie can be used with some preprocessing.
>
> On Thu, Nov 24, 2011 at 11:59 PM, atul anand <atul.87fri...@gmail.com>wrote:
>
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>
> >http://blog.notdot.net/2007/4/Damn-Cool-Algorithms-Part-1-BK-Trees
>
> > this would help.
>
> > On Thu, Nov 24, 2011 at 9:49 PM, Vijay Meena <vijay...@gmail.com> wrote:
>
> >> Can you please elaborate...
>
> >> On Thu, Nov 24, 2011 at 12:14 AM, atul anand 
> >> <atul.87fri...@gmail.com>wrote:
>
> >>> 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
>
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>
>  Regards*
> *"The Coder"*
>
> *"Life is a Game. The more u play, the more u win, the more u win , the
> more successfully u play"*

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