On Mon, Oct 12, 2009 at 1:29 PM, ankur aggarwal ankur.mast@gmail.comwrote:
@shishir
can u give the ds
data structure used is a binary tree, with each node key being the maximum
of its two children's key values.
Space complexity of implementation of tournament principle is O(n).
how would
in that case it is ok..
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*You are given a integer and you want to rotate the bits of the number by a
value x. Consider the right rotation by x means the least significant x bits
should go out from left and take the position of most significant x bits.*
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You received
temp = (0x (32 - x)) n;
n = (n x) | ( temp (32 -x));
On Mon, Oct 12, 2009 at 5:32 PM, ankur aggarwal ankur.mast@gmail.comwrote:
*You are given a integer and you want to rotate the bits of the number by
a value x. Consider the right rotation by x means the least significant x
@raghav
plz give an example
wat r u trying 2 do.
On Mon, Oct 12, 2009 at 8:41 PM, Raghavendra Sharma
raghavendra.vel...@gmail.com wrote:
temp = (0x (32 - x)) n;
n = (n x) | ( temp (32 -x));
On Mon, Oct 12, 2009 at 5:32 PM, ankur aggarwal
ankur.mast@gmail.comwrote:
*You
one question though?
why do you prefer division by bit operations, when there is / operator
( 32 bit div operation is inbuilt in almost all the processors so it
can be done in constant time)
also you recursion is not Tail-recursive so stack will grow like
crazy!!
On Oct 10, 9:36 pm, harit
How abt this..?
for(i=0;ix;i++)
{
res=no1U;
no=no1;
if(res==1)
no=no|32768U;
else
no=no|0U;
}
printf(\nFinal value %u,no);
On Oct 12, 8:11 pm, Raghavendra Sharma raghavendra.vel...@gmail.com
wrote:
temp = (0x (32 - x)) n;
n = (n x) | ( temp (32
@Ankur I am assuming the integer to be 32 bits. actually it should be
0x
step 1 : temp = (0x (32 - x)) n;
step 2 : n = (n x) | ( temp (32 -x));
The first step extracts the lower x bits and second step moves upper bits to
left side and puts the lower x bits at the
