@Johari ,

You have correctly mapped but the question demanded , to remove all
repetition , though the string is still the same as it was given in input .

On Fri, Jun 3, 2011 at 7:00 PM, Kunal Patil <[email protected]> wrote:

> If you are not going to allow  extra space, you have to compromise on time
> complexity..[?]
> If you dont have your string already stored in a trie/hashmap usage of it
> requires additional buffer.
> Simple solution would be:
> Sort given string using in-place sorting algorithm and then removal of
> duplicate characters becomes O(n).
> Total time complexity - O(nlogn) where n --> number of characters in the
> input string.
>
>
> On Thu, Jun 2, 2011 at 11:17 PM, Aakash Johari <[email protected]>wrote:
>
>> It was given that one or two extra variables are allowed. So I used a
>> variable instead for mapping.
>> It is simply mapping of each character in alphabet to a bit in the
>> variable.
>>
>>
>> On Thu, Jun 2, 2011 at 7:10 AM, Ashish Goel <[email protected]> wrote:
>>
>>> using bitmap, but  extra memory not allowed?
>>>
>>>
>>> Best Regards
>>> Ashish Goel
>>> "Think positive and find fuel in failure"
>>> +919985813081
>>> +919966006652
>>>
>>>
>>> On Thu, Jun 2, 2011 at 7:38 PM, Ashish Goel <[email protected]> wrote:
>>>
>>>> what is the logic, kindly explain
>>>> Best Regards
>>>> Ashish Goel
>>>> "Think positive and find fuel in failure"
>>>> +919985813081
>>>> +919966006652
>>>>
>>>>
>>>>
>>>> On Sat, May 28, 2011 at 12:23 PM, Aakash Johari 
>>>> <[email protected]>wrote:
>>>>
>>>>> Following code works for [A-Za-z], can be extended for whole
>>>>> character-set :
>>>>>
>>>>>> #include <stdio.h>
>>>>>>
>>>>>> int main()
>>>>>> {
>>>>>>     unsigned long long int a = 0;
>>>>>>     char str[50];
>>>>>>     int i;
>>>>>>
>>>>>>     scanf ("%s", str);
>>>>>>
>>>>>>     for ( i = 0; str[i]; i++ ) {
>>>>>>         if ( str[i] >= 'A' && str[i] <= 'Z' ) {
>>>>>>             if ( (a & (1ULL << (str[i] - 'A'))) == 0 ) {
>>>>>>                 a |= (1ULL << (str[i] - 'A'));
>>>>>>                 putchar (str[i]);
>>>>>>             }
>>>>>>         } else if ( str[i] >= 'a' && str[i] <= 'z' ) {
>>>>>>             if ( (a & (1ULL << (str[i] - 'a' + 26))) == 0 ) {
>>>>>>                 a |= (1ULL << (str[i] - 'a' + 26));
>>>>>>                 putchar(str[i]);
>>>>>>             }
>>>>>>         }
>>>>>>     }
>>>>>>
>>>>>>     return 0;
>>>>>> }
>>>>>>
>>>>>>
>>>>>>
>>>>> On Fri, May 27, 2011 at 11:15 PM, saurabh singh <
>>>>> [email protected]> wrote:
>>>>>
>>>>>> string getStringWithoutDuplicateChars(string input)
>>>>>> {
>>>>>>
>>>>>> create_empty_trie_ds (say trie)
>>>>>>
>>>>>> integer count = 0;
>>>>>>
>>>>>> for_each_char_in_string (say ch)
>>>>>> {
>>>>>>
>>>>>>     if(trie->contains(ch)) //if ch not there in ds then add it and
>>>>>> return false otherwise return true
>>>>>>     {
>>>>>>          input.remove(count)
>>>>>>      }
>>>>>>
>>>>>>    count++
>>>>>> }
>>>>>>
>>>>>> return input
>>>>>> }
>>>>>>
>>>>>> On Sat, May 28, 2011 at 11:32 AM, Rajeev Kumar <
>>>>>> [email protected]> wrote:
>>>>>>
>>>>>>> Design an algorithm and write code to remove the duplicate characters
>>>>>>> in a string without using any additional buffer.
>>>>>>>  NOTE: One or two additional variables are fine.
>>>>>>>  An extra copy of the array is not.
>>>>>>>
>>>>>>>
>>>>>>> --
>>>>>>> Thank You
>>>>>>> Rajeev Kumar
>>>>>>>
>>>>>>> --
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>>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>> --
>>>>>> Thanks & Regards,
>>>>>> Saurabh
>>>>>>
>>>>>> --
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>>>>>>
>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> -Aakash Johari
>>>>> (IIIT Allahabad)
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>  --
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>>>>>
>>>>
>>>>
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>>
>>
>>
>> --
>> -Aakash Johari
>> (IIIT Allahabad)
>>
>>
>>
>>
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>
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-- 
Lalit Kishore Sharma,

IIIT Allahabad (Amethi Capmus),
6th Sem.

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