@anurag : there is no need to SORT. as it will increase complexity O(n) to
O(n log n).
here is the correct code.. please look over it and notify me if i'm wrong .
T.C. = O( n )
// ex: "1 4 3 2 0" i'm explaining on behalf of it.
bool permute (int *arr , int N )
{
if ( N<=1 ) return false;
for ( int i = N-2 ; i >= 0 ; i-- )
{
if ( arr[i+1] > arr[i]) // now i points to 1
{
for ( int j = N-1 ; j > i ; j--)
{
if ( arr[i]< arr[j]) // now j points to 2(just greater than
1 )
{
swap(arr[i],arr[j]) ;//this will generate "2 4 3 1 0"
since 4310 are already sorted in reverse
//thus no need to sort again in
inc order rather than reversing
i++ ; j = N-1;
while(i<j) swap(arr[i++],arr[j--]) ; // reversing the
seq 4..0 to 0..4
return true ;
}
}
}
}
return false ;
}
Regards,
Ritesh Kumar Mishra
Information Technology
Third Year Undergraduate
MNNIT Allahabad
On Mon, Dec 24, 2012 at 7:56 PM, marti <amritsa...@gmail.com> wrote:
>
> I REPEAT, THERE is no need to SORT;
>
>
> http://en.wikipedia.org/wiki/Permutation#Lexicographical%5Forder%5Fgeneration
>
>
> On Friday, December 14, 2012 11:56:16 AM UTC+5:30, tapan rathi wrote:
>
>> For a given number, find the next greatest number which is just greater
>> than previous one and made up of same digits.
>
> --
>
>
>
--
