@Rajeev: The above algorithm assumes a source string and a destination
string. But here you are provided only the source string. And you will have
to edit it (minimally) such that the resulting string matches a word in the
dictionary.

Need slight modification. Looking for the modification. (Interviewer told.
The same answer was given).

On Mon, Nov 14, 2011 at 4:59 AM, rajeev bharshetty <[email protected]>wrote:

> 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.
> > >>
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