@mohit: that will modify the original array
and also time =O(nlogn)...
On Mon, Sep 5, 2011 at 1:01 AM, Ankuj Gupta wrote:
> @mohit: that will modify the original array
>
> On Sep 4, 6:40 pm, sarath prasath wrote:
> > here is my approach
> > where i left the non repeating characters as it is and
@mohit :good one
On Sun, Sep 4, 2011 at 4:34 PM, mohit verma wrote:
> hey guys
> why hashing?
> can't we simply sort the array of characters in-place - space complexity
> O(1) .
> and then simply rearrange the array in-place with character+frequency.
>
> On Sun, Sep 4, 2011 at 3:13 PM, bharatkum
hey guys
why hashing?
can't we simply sort the array of characters in-place - space complexity
O(1) .
and then simply rearrange the array in-place with character+frequency.
On Sun, Sep 4, 2011 at 3:13 PM, bharatkumar bagana <
bagana.bharatku...@gmail.com> wrote:
> +1 rahul and +1 ankuj
>
>
> On S
+1 rahul and +1 ankuj
On Sat, Sep 3, 2011 at 11:37 AM, icy` wrote:
> Just use a hash to count frequency of something; e.g. in ruby:
> ar= %w(a a b c a b b c d e a d e f)
> freq=Hash.new(0)
> ar.each {|c| freq[c]+=1}
> p freq
>
> #output
> #{"a"=>4, "b"=>3, "c"=>2, "d"=>2, "e"=>2, "f"=>1}
>
> yo
@ankuj just want to clarify that in hashing method we require array of
fixed size let say arr[26] , so is it considered as constant space or not?
On Sat, Sep 3, 2011 at 8:02 PM, siddharam suresh wrote:
> sol already posted please search old thread
> Thank you,
> Sid.
>
>
>
> On Sat, Sep 3, 2011
sol already posted please search old thread
Thank you,
Sid.
On Sat, Sep 3, 2011 at 8:01 PM, Ankuj Gupta wrote:
> If we take our input to be characters a-z ans A-Z then we require
> fixed space which can be treated as O(1).
> On Sep 3, 7:10 pm, teja bala wrote:
> > this 'll work if u i/p the s
