tell us the logic...
ankur
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If the set has fewer elements than an integer has bits, just count
from 1 to MAXINT. If bit i is 0, the element is not in the set, and if
bit i is 1, the element is in the set.
Dave
On Aug 26, 2:20 pm, AKS abhijeet.k.s...@gmail.com wrote:
Hello fellas,
i am lookin for an algorithm to find
Please see http://geeksforgeeks.org/?p=588
Well explained and coded solution.
On Thu, Aug 27, 2009 at 5:28 AM, Dave dave_and_da...@juno.com wrote:
If the set has fewer elements than an integer has bits, just count
from 1 to MAXINT. If bit i is 0, the element is not in the set, and if
bit i
The following will generate an output like this - 0
0 1
0 1 2
0 1 2 3
0 1 3
0 2 3
0 3
1
1 2
1 2 3
1 3
2
2 3
3
using these indices you can generate from any given set.
class SetGenerator
{
public:
SetGenerator(size_t length)
: LENGTH(length)
{
can i have the algorithm only in plain simple english
rather than havin the whole code ???
it ll be really helpful if u tell me the logic
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You just gt generate this pattern of indices into the set -
0
0 1
0 1 2
0 1 2 3
0 1 3
0 2 3
0 3
1
1 2
1 2 3
1 3
2
2 3
3
just figure out the conditions to generate the above yourself ... and you'll
figure out what the code does
On Wed, Aug 26, 2009 at 10:01 PM, Abhijeet Kumar
