int main()
{
char *s="nitan";
int n,i,j,c=0;
char *d;
n=strlen(s)/2;
//printf("%d",n);
for(i=1;i<=n;i++)
{
if(s[n-i]!=s[n+i])
break;
}
d=strcpy(d,s);
for(j=i;j<=n;j++)
{
if(d[n+j]!=d[n-j])
{
c++;
d[n+j]=d[n-j];
}
}
printf("%s %d",d,c);
}
On 6 September 2011 01:21, Pratz mary <[email protected]> wrote:
> for yahoo shudnt min additions be yahay?
>
>
> On 6 September 2011 00:48, learner <[email protected]> wrote:
>
>> @sandeep Explain Algorithm/logic & Time Complexity ?
>>
>> Thanks
>>
>> --
>> You received this message because you are subscribed to the Google Groups
>> "Algorithm Geeks" group.
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>>
>>
>
>
> --
> regards Pratima :)
>
--
regards Pratima :)
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