formatting changes .....
void permute(char *a, int i, int n)
{
int j;
if (i == n)
printf("%s\n", a);
for (j = i; j <= n; j++)
{
if(a[i] != a[j]) ////check before swapping if you are
swapping different elements or not, if not then don't.
{ swap(a[i], a[j]);
permute(a, i+1, n);
swap([ai], a[j]);
}
}
}
-----------------
....
On Wed, Mar 12, 2014 at 11:59 AM, navneet singh gaur
<navneet.singhg...@gmail.com> wrote:
> void permute(char *a, int i, int n)
> {
> int j;
> if (i == n)
> printf("%s\n", a);
>
> for (j = i; j <= n; j++)
> {
> if(a[i] != a[j])
> { swap(a[i], a[j]); //just check before swapping if
> you are swapping different elements
> //or not, if not then don't swap
> permute(a, i+1, n);
> swap([ai], a[j]);
> }
> }
> }
>
>
> On Sat, Jan 11, 2014 at 4:09 PM, bujji jajala <jajalabu...@gmail.com> wrote:
>> Hi Don,
>> Good one :) Nice to see different approaches for this problem.
>>
>> -Thanks,
>> Bujji
>>
>>
>> On Fri, Jan 10, 2014 at 9:11 AM, Don <dondod...@gmail.com> wrote:
>>>
>>> Sort the input string and remove duplicates, keeping a count of the number
>>> of occurrences of each character.
>>> They you can build the permutations easily.
>>>
>>> For your example, you would start with
>>>
>>> char *str = "aba";
>>> int len = strlen(str);
>>>
>>> Which would be converted to
>>>
>>> char *str "ab";
>>> int rep[N] = {2,1,0}; // The string contained 2 'a's and 1 'b'
>>> char result[N];
>>>
>>> Then call permute(str,rep,len)
>>>
>>> void permute(char *str, int *rep, int len, int p=0)
>>> {
>>> if (p<len)
>>> {
>>> for(int i = 0; str[i]; ++i)
>>> if (rep[i])
>>> {
>>> result[p] = str[i];
>>> --rep[i];
>>> permute(str, rep, len,p+1);
>>> ++rep[i];
>>> }
>>> }
>>> else printf("%s\n", result);
>>>
>>> }
>>>
>>> On Monday, January 6, 2014 5:05:08 PM UTC-5, bujji wrote:
>>>>
>>>> generate all possible DISTINCT permutations of a given string with some
>>>> possible repeated characters. Use as minimal memory as possible.
>>>>
>>>> if given string contains n characters in total with m < n distinct
>>>> characters each occuring n_1, n_2, ....n_m times where n_1 + n_2 + ...+ n_m
>>>> = n
>>>>
>>>> program should generate n! / ( n_1! * n_2! * ....* n_m! ) strings.
>>>>
>>>> Ex:
>>>> aba is given string
>>>>
>>>> Output:
>>>>
>>>> aab
>>>> aba
>>>> baa
>>>>
>>>>
>>>> -Thanks,
>>>> Bujji
>>>>
>>> --
>>> You received this message because you are subscribed to the Google Groups
>>> "Algorithm Geeks" group.
>>> To unsubscribe from this group and stop receiving emails from it, send an
>>> email to algogeeks+unsubscr...@googlegroups.com.
>>
>>
>> --
>> You received this message because you are subscribed to the Google Groups
>> "Algorithm Geeks" group.
>> To unsubscribe from this group and stop receiving emails from it, send an
>> email to algogeeks+unsubscr...@googlegroups.com.
>
>
>
> --
> navneet singh gaur
--
navneet singh gaur
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to algogeeks+unsubscr...@googlegroups.com.
