It can be done inplace, then Time Complexity will be O( N^2 x W ) where N
is no. of words and W is word size.
Start from the left and sort the current word.
Keep a pointer PTR to the first index of the element left to process.
Initially it will be 0.As we add words this pointer moves forwards.
Now iterate the whole list of words to find all those words having same
sorted sequence i.e. anagrams
Whenver we get a word which is anagram of the current string, swap it with
the string pointed by PTR, move PTR forward.
pseudoCode :-
func( vector<string> v)
{
int n=v.size();
for(int i=0;i<n;i++)
{
string currentWord = v[i];
sort(currentWord);
for(int j=i+1;j<n;j++)
{
if ( sort(v[j]) == currentWord ) // anagrams found,
{
swap( v[i] , v[j] ); //move this string to
the position next to pos of currentWord
i++; //and move the index
of currentWord forward
}
}
}
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To view this discussion on the web visit
https://groups.google.com/d/msg/algogeeks/-/h04_lKqCILQJ.
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.
