@sunny

the no of set bits in m will tell what all length(4,2 in above case)
are need to be merged.
e.g if if  m ==6 then m = 0110
 since bit set position are 2 and 1.
so length of rope  need to combine is 2^2=4 and 2^1 = 2;i.e 4 and 2

Thnaks
Santosh

On Sat, Jul 2, 2011 at 2:58 PM, santosh mahto <santoshbit2...@gmail.com>wrote:

> @sunny
>
> that will  work fine(xoring).
> In place of Xoring u can also do OR of two number and find the distance
> between fist set bit from left and first set bit from right,
>
> Since bit operation is really fast operation so best algo this is of
> complexity O(1);
>
> Explanation How it works:
>
> In l only one bit will be set. In m  multiple bit  would be set with all
> bit before the bit set in l.
> NOW suppose u divde l in 2 parts means u have shifted set bit of l to the
> right.
> dividing again will shift set bit in l to the right.
> u have to divide till  set bit in l reached the position where first  bit
> of m is set.
>
> how many times u have shifted is count of  dividng the rope
>
> its Simple.
>
> u can check with any example
>
>
> Thanks
> Santosh
>   On Sat, Jul 2, 2011 at 2:41 PM, sunny agrawal 
> <sunny816.i...@gmail.com>wrote:
>
>> for a number N
>> first set bit(From Left) is simply integer value of log(N)
>> last set bit can be calculated as
>>
>> N = N-(N&(N-1)); and then Log(N)
>>
>> int i = log(n);
>> n -= n&(n-1);
>> int j = log(n);
>>
>> i-j will be the answer.
>>
>>
>>
>> On Sat, Jul 2, 2011 at 2:34 PM, cegprakash <cegprak...@gmail.com> wrote:
>>
>>> oh fine.. got it now.. set bit is '1' right.. and is there any short
>>> ways to find the difference between first set and short set bit
>>> without dividing by 2 repeatedly?
>>>
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>>
>>
>> --
>> Sunny Aggrawal
>> B-Tech IV year,CSI
>> Indian Institute Of Technology,Roorkee
>>
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