@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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