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
<[email protected]> 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 <[email protected]> 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 <[email protected]> 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
>>>>
>>> --
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>>> email to [email protected].
>>
>>
>> --
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>> email to [email protected].
>
>
>
> --
> navneet singh gaur
--
navneet singh gaur
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