@dave: actually how did u get the approach to this? I mean why did u have to
do the q|=1 in the if(a>=b) condition and q<<=1 always in the loop?

On Fri, Aug 19, 2011 at 1:46 PM, Sanjay Rajpal <[email protected]>wrote:

> @Shashank : Would you throw some light on how you determined the complexity
> ?
>
>
> *Regards
>
> Sanju
>
> Happy to Help :)*
>
>
>
> On Fri, Aug 19, 2011 at 2:36 AM, WgpShashank 
> <[email protected]>wrote:
>
>> We can use bit logic to reduce the time complexity to O(logn) where n is
>> Quotient
>>
>> Algorithm will be as follow as we know
>>
>> 1. Left shifting an unsigned number by 1 multiplies that number by 2.
>> 2. Right shifting an unsigned number by 1 divides that number by 2.
>>
>> Therefore the procedure for the division algorithm, given a dividend and a
>> divisor .
>> core logic will be to left shift (multiply by 2) untill its greater then
>> dividend , then continue this routine with the the difference between the
>> dividend and divisor and divisor till the point where dividend is less than
>> divisor or their difference is zero.
>>
>> Lets see one example: dividend=23 divisor=3
>>
>> then left shift for 3 i.e 3 will converted to3, 6,12,24,… since 12 <23
>> hence now dividend is 11 and quotient in 4(two time shift operation) now
>> again 3,6,12.. 6<11 hence dividend is 11-6=5 and quotient =4+2=6 now only
>> 3<5 hence remainder =2 quotient =6+1=7 so answer.
>>
>> Time Complexity O(logn) Number of bits in Quotient
>>
>> Correct me if anything wrong
>>
>> *Thanks
>> Shashank Mani
>> Computer Science
>> Birla Institute of Technology Mesra*
>>
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