#include<conio.h>
#include<stdio.h>
#include<string.h>
#include<iostream.h>
void substr(char *dst,char *src, size_t start, size_t stop)
{
int count = stop - start;
sprintf(dst, "%.*s", count, src + start);
}
int ispalin(char *k)
{
int i1=0;
int i2=strlen(k)-1;
while(i2>i1)
{
if(k[i2]!=k[i1])
return 0;
i1++;
i2--;
}
return 1;
}
void makepalin(char *t)
{
char s[50];
char *p=t;
int i=0;
char k[50];
while (!ispalin(p))
{
s[i]=p[0];
substr(p,p,1,(strlen(t)));
i++;
}
strcpy(k,s);
strcat(k,p);
strcat(k,strrev(s));
printf(" palin is %s ",k);
}
int main( )
{
char s[50];
gets(s);
makepalin(s);
getch();
}
On Mon, Sep 5, 2011 at 10:46 AM, hemank lamba <[email protected]> wrote:
> @Sukran: This wont work for the test case like this
>
> for example the word is "Nitan":
> then the word ur algorithm will create is "Nitanatin" hence the number of
> additions =4
>
> but ideal case i would be
> "nitatin" : where number of additions is only 2.
>
> On Sun, Sep 4, 2011 at 11:11 PM, sukran dhawan <[email protected]>wrote:
>
>>
>>
>> On Sun, Sep 4, 2011 at 11:11 PM, sukran dhawan <[email protected]>wrote:
>>
>>> for(i0;i<n/2;i++)
>>> {
>>> a[n-1-i] = a[i];
>>> }
>>>
>>> will this work ?
>>> where n is the length os string
>>>
>>>
>>> On Sun, Sep 4, 2011 at 7:54 PM, learner <[email protected]>wrote:
>>>
>>>> Given a word, convert it into a palindrome with minimum addition of
>>>> letters to it. letters can be added anywhere in the word. for eg if
>>>> yahoo is given result shud be yahohay.
>>>>
>>>>
>>>> Thanks
>>>>
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>>>>
>>>
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>
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