This code works perfectly. Try out with different test data.
int main()
{
//char str[]="qwertyuioplkjhgfdsazxcvbnm";
char str2[] = "qwertyquejolpd";
int val, i;
int len = strlen(str);
int unique=1;
int checker=0;
for(=0;i<len;i++)
{
val = str[i]-'a';
if( (checker & (1<<val)) > 0 )
{
printf("\nNo, there aren't unique.");
unique=0;
break;
}
checker |= (1<<val);
}
if(unique==1)
printf("\nUnique charaters.");
return 1;
}
On Sat, Jul 23, 2011 at 12:18 PM, rajeev bharshetty <[email protected]>wrote:
> @Victor : Counting sort uses extra data structures to maintain the count
> and the output array , so it is not viable I think ?? what say ??
>
>
> On Sat, Jul 23, 2011 at 11:04 AM, Victor Manuel Grijalva Altamirano <
> [email protected]> wrote:
>
>> Use a type of counting sort with ascii
>>
>>
>> 2011/7/22 SkRiPt KiDdIe <[email protected]>
>>
>>> Use 8 mask integers or 4 LL to cover the complete ascii range.
>>>
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>>
>>
>>
>> --
>> Victor Manuel Grijalva Altamirano
>> Universidad Tecnologica de La Mixteca
>>
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>
>
>
> --
> Regards
> Rajeev N B <http://www.opensourcemania.co.cc>
>
>
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Regards
Reynald Reni
Masters in Software Engineering
CIT - India
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