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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>
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