use hash table.characters count will be stored on the basis of their ascii
value . ie a's count will be stored in array[97] .
at last just check the count of each . count==1 gives u the first non
repeating character
On Tue, Jul 19, 2011 at 12:31 AM, aditi garg <aditi.garg.6...@gmail.com>wrote:
> @sagar ur solution is very gud...bt wat abt blank spaces and capital
> alphabets dat may be thr...i dnt think it will take care of that...
>
>
> On Mon, Jul 18, 2011 at 11:32 PM, varun pahwa <varunpahwa2...@gmail.com>wrote:
>
>> @sagar: that's what i have done i have taken two variables x and y which
>> can show if repetition is there or not. and we can store the initial
>> positions in the array in that case u don't have to traverse the string
>> twice.
>>
>>
>> On Mon, Jul 18, 2011 at 10:33 PM, sagar pareek <sagarpar...@gmail.com>wrote:
>>
>>> @dumanshu
>>> first read my soution just above yours :)
>>>
>>> On Mon, Jul 18, 2011 at 10:21 PM, Dumanshu <duman...@gmail.com> wrote:
>>>
>>>> heres my solution with TC O(n) and SC O(26)
>>>> input string str
>>>> int arr[26] = {0};
>>>> traverse the string, character by character and increment the
>>>> corresponding counter.
>>>> i.e. arr[str[i]]++;
>>>>
>>>> Now traverse the string again and print out the first character
>>>> encountered whose arr[str[i]] == 1;
>>>>
>>>> On Jul 18, 9:20 pm, sagar pareek <sagarpar...@gmail.com> wrote:
>>>> > Very good solution :- but space complexity = O(26)
>>>> >
>>>> > take integer array arr[0-25] and initialise it with 0 by taking it
>>>> static
>>>> > logic is that we have only 26 characters so if i want to map character
>>>> 'a'
>>>> > with 0th position of arr[] then it can be done as atoi('a')-97.
>>>> > so whenever we encounter any character say str[i] (where str is array
>>>> of
>>>> > given string) then it can be incremented as arr[atoi(str[i])-97]++
>>>> > so traverse the whole str[] and increment the corresponding values .
>>>> > At the end those characters which never encounter have values 0 in arr
>>>> ,
>>>> > which encounter only once have values 1 and more than once have
>>>> values>1.
>>>> > at the end traverse the whole arr[] and find out the corresponding
>>>> character
>>>> > as itoa(arr[i]+97) :) :)
>>>> >
>>>> > But we have to do extra work to find the first character which repeats
>>>> only
>>>> > once
>>>> >
>>>> > On Mon, Jul 18, 2011 at 8:09 PM, hary rathor <harry.rat...@gmail.com>
>>>> wrote:
>>>> > > can we use bit vector ?
>>>> > > because by do it we need just 32 bits of one extra variable .
>>>> >
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>>>> > **Regards
>>>> > SAGAR PAREEK
>>>> > COMPUTER SCIENCE AND ENGINEERING
>>>> > NIT ALLAHABAD
>>>>
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>>>
>>>
>>> --
>>> **Regards
>>> SAGAR PAREEK
>>> COMPUTER SCIENCE AND ENGINEERING
>>> NIT ALLAHABAD
>>>
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>>
>>
>>
>> --
>> Varun Pahwa
>> B.Tech (IT)
>> 7th Sem.
>> Indian Institute of Information Technology Allahabad.
>> Ph : 09793899112
>> Official Email :: rit2008...@iiita.ac.in
>> Another Email :: varunpahwa.ii...@gmail.com
>>
>> People who fail to plan are those who plan to fail.
>>
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>
>
>
> --
> Aditi Garg
> Undergraduate Student
> Electronics & Communication Divison
> NETAJI SUBHAS INSTITUTE OF TECHNOLOGY
> Sector 3, Dwarka
> New Delhi
>
> 9718388816
>
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