Assuming string as only lower case characters.
now declare two unsigned int as x,y and ar[26].
now when u encounter any character c among (a,...z) first time u will set
(c-'a') bit of x and check if it has already been occurred then set (c-'a')
bit of y.and ar[i] will be storing the very first occurrence of the c
character where i will be (c-'a').
now traverse the array and output will be (i+'a') where a[i] is min and
(ith) bit in x is set and ith bit in y bit is unset.
On Mon, Jul 18, 2011 at 7:26 PM, vaibhav shukla <vaibhav200...@gmail.com>wrote:
> one solution could be:
>
> #include<stdio.h>
> #include<string.h>
> char non_repetition(char *p,int size)
> {
> int i,j,flag=0;
> for(i=0;i<size;i++)
> {
> for(j=0;j<size;j++)
> {
> if(j==i)
> continue;
> if(p[i]==p[j])
> flag=1;
> }
> if(flag==0)
> return p[i];
> else
> flag=0;
> }
> return '\0';
> }
>
> int main()
> {
> char
> str[]="aaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbccccccccccccccccccccccc
> *V*cccccccccccccccccccccccc";
> int len=strlen(str);
> char firstNR=non_repetition(str,len);
> if(firstNR=='\0')
> printf("all are repeating at least once\n");
> else
> printf("first non repeated character is=%c\n",firstNR);
> return 0;
> }
>
> gives *V* as the output
>
> On Mon, Jul 18, 2011 at 6:45 PM, Dumanshu <duman...@gmail.com> wrote:
>
>> You are given a long string and you have to print the "first" non
>> repeating character.
>> Solve it keeping SC and TC in mind. That is present both solutions,
>> one with high SC and low TC and viceversa.
>>
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>>
>
>
> --
> best wishes!!
> Vaibhav
> DU-MCA
>
>
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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
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