i agree.. it is not the the fastest algorithm for this problem.. it is a
basic solution..

On Sun, Jul 10, 2011 at 2:09 PM, Anurag Aggarwal <[email protected]
> wrote:

> Sunny
> Don.t you think your method is very slow as you are checking every number
> that for number of set bits and if it is a equal to desired than you are
> decreasing n i.e. required number.
> Even if when n=1 and k=32 your solution will start with 0 and go up to
> 2^31 but the answer could be found in single iteration.
>
> So is there any way to do it a little faster.
>
> On Sun, Jul 10, 2011 at 1:08 PM, Sunny T <[email protected]> wrote:
>
>> sorry ankit... ur solution works only for.. k=1 and n=3...
>> try for k=2 and n=6.. then the output should be 12...
>> similarly for k=3 n=1.. the output should be...7...
>>
>> so plz correct ur code..
>>
>>
>> On Sun, Jul 10, 2011 at 12:38 PM, ankit sambyal 
>> <[email protected]>wrote:
>>
>>> Here is my approach :
>>>
>>> int main()
>>> {
>>>    int a=1,k=1,n=3;
>>>    while(k>1)
>>>    {
>>>        k--;
>>>        a=(a<<1) | 1;
>>>    }
>>>    while(n>1)
>>>    {
>>>        a=a<<1;
>>>        n--;
>>>    }
>>>    printf("%d",a);
>>>    return 0;
>>> }
>>>
>>> On Sat, Jul 9, 2011 at 11:14 PM, anurag <[email protected]>
>>> wrote:
>>> > You are given two integers n and k
>>> > k signifies number of set bits i.e. if k = 3 then output should have 3
>>> > set bits.
>>> > Output should be the nth smallest number having k set bits
>>> > for example
>>> > k=1 and n=3
>>> > output should be
>>> > 4 (00000100)
>>> >
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>>
>>
>> --
>> Warm Regards,
>> Sunny T
>>
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-- 
Warm Regards,
Sunny T

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