OK, I see what you mean. However, I have a feeling that this way we
don't actually get a constant storage - that is,
it is 'constant' in a sense that it doesn't depend on the number of
integers in array, but rather on the maximum element of the array. I'm
not sure that such a solution would be correct for the original
problem, but I'll nevertheless propose it to the author.
Thanks for the idea!
On Aug 22, 6:13 am, L7 <[EMAIL PROTECTED]> wrote:
> Sorry. I wrote too quickly. The amount you need is not determined by
> log, but by division.
> For the sake of argument, say you could hold the value 0 - 255. This
> means we are dealing with 8-bit storage, but it makes the example
> easier. You would need, 255/8 or roughly 32 distinct integers to do
> it. That number wont ever change. Regardless if you have 2 numbers or
> 257 numbers in your array. If you scale that to 32 bit numbers, you
> get the following requirement ~ 135,000,000 distinct integers. It is a
> large requirement, but it is constant and capable of determining what
> you want regardless if it has 2 or 4,294,967,297 elements.
>
> Now, if you want to minimize that storage, you can scan the array
> first and pick out the maximum of the array. That is an O(n)
> operation. Base your static 'hash' on that maximum and you use the
> smallest amount of space. Realize that O(n) + O(n) results in O(n).
>
> On Aug 21, 1:08 am, dsha <[EMAIL PROTECTED]> wrote:
>
> > Sorry, my own calculations are wrong - 7 32-bit integers can only
> > represent 32 * 7 bits of information,
> > I mistakenly added the redundant term '8' in the previous post. But
> > the idea is the same - 32 * 7 bits of information
> > cannot hold presence indicators for 2^32 numbers.
>
> > On Aug 21, 9:06 am, dsha <[EMAIL PROTECTED]> wrote:
>
> > > Let's count: log base 32 of the max value of an integer, or, rather,
> > > total amount of integers which can be stored in 32 bits (=2^32), is
> > > equal to
> > > ceil(log_32(2^32)) = 7. But 7 32-bit integers can only represent 32 *
> > > 7 * 8 bits of information
> > > which is not enough to store membership information for arbitrary
> > > numbers in range [1...2^32]. Sorry if I've again
> > > missed something in your proposal...
>
> > > On Aug 21, 6:14 am, L7 <[EMAIL PROTECTED]> wrote:
>
> > > > On Aug 20, 3:10 pm, dsha <[EMAIL PROTECTED]> wrote:
>
> > > > > Sorry, it's too late here, so I'm not sure I've completely understood
> > > > > the solution... Am I right that what is proposed
> > > > > is to create a huge bit vector of size (if I'm not wrong) 2^32 bits
>
> > > > Not exactly. You can store 32 unique indicies in a 32-bit integer. So
> > > > basically you need log base 32 of the max value of an integer.
>
> > > > > (assuming 32-bit integers) and use the value of an element array as an
> > > > > index to this vector? Then a linear algorithm solving the original
> > > > > problem is indeed not hard to figure out... The only problem is that
> > > > > I'm not sure if the array of size of 2^32 / 8 bytes can be considered
> > > > > a constant memory...
>
> > > > If you only speak of 32-bits (or 64, whichever the case) then you will
> > > > never have to resize the array at any time. Indeed, you will have a
> > > > large amount of unused space (for smaller arrays) - but the size does
> > > > not change with the size of the array and therefore can be considered
> > > > constant.
>
> > > > > On Aug 20, 1:10 am, L7 <[EMAIL PROTECTED]> wrote:
>
> > > > > > This can be done with constant space and O(n) time if you hold to
> > > > > > _exactly_ what is mentioned in your post. An integer (32/64) can
> > > > > > hold
> > > > > > only a finite number of values. Based on that, you can create a
> > > > > > bitfield at size log of that mamimum value. You now have constant
> > > > > > size
> > > > > > 'hashing' where the hash is simply 1 << x[n]. Now the hash is not a
> > > > > > problem of growth based on the size of the array. With that
> > > > > > implementation of sumedh sakdeo's hash solution you have the problem
> > > > > > solved.
>
> > > > > > On Aug 19, 1:45 am, dsha <[EMAIL PROTECTED]> wrote:
>
> > > > > > > Hello,
>
> > > > > > > as I already mentioned, there are no any additional constraints
> > > > > > > on the
> > > > > > > input data, so you cannot assume
> > > > > > > that the array is ordered. Nor can you order it, as I presume
> > > > > > > that the
> > > > > > > input data should remain intact (sorry, I must have mentioned this
> > > > > > > before). So a solution cannot be based upon the assumption you
> > > > > > > made.
>
> > > > > > > Thanks.
>
> > > > > > > On Aug 18, 2:14 pm, "Peeyush Bishnoi" <[EMAIL PROTECTED]>
> > > > > > > wrote:
>
> > > > > > > > Hello All
> > > > > > > > I thanxs Vaibhav to give feedback on my solution....
>
> > > > > > > > I am once again putting my solution back on this group. I
> > > > > > > > welcome ur all
> > > > > > > > valuale feedback on this...
> > > > > > > > I can think this solution will work with constant space &
> > > > > > > > linear time ....
> > > > > > > > incomparison to my previous solution which is n*n at worst case.
> > > > > > > > But this solution need integer array data set to be sorted...
>
> > > > > > > > int main(){
> > > > > > > > int a[]={1,2,2,3};
> > > > > > > > int count=sizeof(a)/sizeof(a[0]);
> > > > > > > > printf("No.of elemenst:%d\n",count);
> > > > > > > > fun(a,count);
> > > > > > > > return 0;
>
> > > > > > > > }
>
> > > > > > > > fun(int a[],int count){
> > > > > > > > int i,j;
> > > > > > > > for(i=0,j=i+1;i<count,(i<j &&
> > > > > > > > (a[i]!=a[j]));i++,j++);
> > > > > > > > if((i<j)&& (a[i]==a[j])){
> > > > > > > > printf("No. is
> > > > > > > > repeated:%d\n",a[i]);
> > > > > > > > }else{
>
> > > > > > > > }
>
> > > > > > > > }
>
> > > > > > > > ---
> > > > > > > > Peeyush Bishnoi
>
> > > > > > > > On 8/18/07, Vaibhav Jain <[EMAIL PROTECTED]> wrote:
>
> > > > > > > > > peeyush,
>
> > > > > > > > > take eg: n=6
> > > > > > > > > array values: 10 20 30 40 50 50
> > > > > > > > > in worst case, while loop which can increment 'i' can go upto
> > > > > > > > > n-1
> > > > > > > > > and for loop (for 'j') every n-1 time check upto n times
> > > > > > > > > so total it becomes (n-1)*n= O(n*n).
>
> > > > > > > > > like this i think u can observe now..
>
> > > > > > > > > On 8/18/07, Peeyush Bishnoi <[EMAIL PROTECTED]> wrote:
>
> > > > > > > > > > Thanxs for giving feedback... :-)
> > > > > > > > > > Can you please explain how worst case time complexity is
> > > > > > > > > > O(n*n) of
> > > > > > > > > > this solution. Means how u determine this. Plz explain....
>
> > > > > > > > > > ---
> > > > > > > > > > Peeyush Bishnoi
>
> > > > > > > > > > On 8/18/07, Vaibhav Jain <[EMAIL PROTECTED]> wrote:
>
> > > > > > > > > > > hi peeyush,
>
> > > > > > > > > > > ur solution is nice, it is brute force method and space
> > > > > > > > > > > complexity is
> > > > > > > > > > > constant here
> > > > > > > > > > > but ur solution contains worst case time complexity O(n*n)
> > > > > > > > > > > and we want O(n) solution. So ur solution is not required
> > > > > > > > > > > solution.
>
> > > > > > > > > > > On 8/18/07, Peeyush Bishnoi < [EMAIL PROTECTED]> wrote:
>
> > > > > > > > > > > > Hello All ,
>
> > > > > > > > > > > > I am thinking this solution offered by me is some what
> > > > > > > > > > > > accurate with
> > > > > > > > > > > > constant space . Just put ur feed back on this.
>
> > > > > > > > > > > > If you have any query ask me.
>
> > > > > > > > > > > > int main(){
> > > > > > > > > > > > int a[]={1,2,2,3};
> > > > > > > > > > > > int count=sizeof(a)/sizeof(a[0]);
> > > > > > > > > > > > printf("No.of elemenst:%d\n",count);
> > > > > > > > > > > > fun(a,count);
> > > > > > > > > > > > return 0;
> > > > > > > > > > > > }
>
> > > > > > > > > > > > fun(int a[],int count){
> > > > > > > > > > > > int i=0,j;
> > > > > > > > > > > > while(i<count){
> > > > > > > > > > > > for(j=0;(j<i && (a[i]!=a[j]));j++);
> > > > > > > > > > > > if((j<i)&& (a[i]==a[j])){
> > > > > > > > > > > > printf("No. is
> > > > > > > > > > > > repeated:%d\n",a[i]);
> > > > > > > > > > > > }else{
> > > > > > > > > > > > }
> > > > > > > > > > > > i++;
> > > > > > > > > > > > }
> > > > > > > > > > > > }
>
> > > > > > > > > > > > ---
> > > > > > > > > > > > Peeyush Bishnoi
>
> > > > > > > > > > > > On 8/18/07, Dondi Imperial < [EMAIL PROTECTED] > wrote:
>
> > > > > > > > > > > > > hi,
>
> > > > > > > > > > > > > actually in mine space complexity is O(n) in _all_
> > > > > > > > > > > > > cases. :). Out
> > > > > > > > > > > > > of
> > > > > > > > > > > > > curiousity, will your solution work when not all the
> > > > > > > > > > > > > numbers in
> > > > > > > > > > > > > the
> > > > > > > > > > > > > range are present in the array?
>
> > > > > > > > > > > > > Thanks,
>
> > > > > > > > > > > > > Dondi
>
> > > > > > > > > > > > > On 8/17/07, Vaibhav Jain < [EMAIL PROTECTED] > wrote:
> > > > > > > > > > > > > > hello Dondi,
>
> > > > > > > > > > > > > > in ur solution, space complexity will be O(n) in
> > > > > > > > > > > > > > worst case.
> > > > > > > > > > > > > > but in my solution it will constant space with
> > > > > > > > > > > > > > linear
> > > > > > > > > > > > > complexity.
>
> > > > > > > > > > > > > > now think abt how to prove it if range is not known
> > > > > > > > > > > > > > for numbers
> > > > > > > > > > > > > > then can we achieve it or not?
> > > > > > > > > > > > > > if not then prove it....???
>
> > > > > > > > > > > > > > On 8/17/07, Dondi Imperial < [EMAIL PROTECTED]>
> > > > > > > > > > > > > > wrote:
>
> > > > > > > > > > > > > > > if you know the range of the numbers don't you
> > > > > > > > > > > > > > > just have to
> > > > > > > > > > > > > create and
> > > > > > > > > > > > > > > array (of length k in your example) then iterate
> > > > > > > > > > > > > > > over the
> > > > > > > > > > > > > array and
> > > > > > > > > > > > > > > increment the corresponding element in the other
> > > > > > > > > > > > > > > array.
>
> > > > > > > > > > > > > > > Ie,
>
> > > > > > > > > > > > > > > int[] arrayValues = some array of a known range
> > > > > > > > > > > > > > > int[] arrayLookup = int[min_in_range -
> > > > > > > > > > > > > > > max_in_range + 1]
>
> > > > > > > > > > > > > > > foreach(i in arrayValues)
> > > > > > > > > > > > > > > if(arrayLookup[i] > 0) then
> > > > > > > > > > > > > > > found
> > > > > > > > > > > > > > > else
> > > > > > > > > > > > > > > arrayLookup[i]++
>
> > > > > > > > > > > > > > > Of course range could be prohibitively large
> > > > > > > > > > > > > > > (still constant
> > > > > > > > > > > > > though if
> > > > > > > > > > > > > > > you know the range before hand).
>
> > > > > > > > > > > > > > > On 8/17/07, Vaibhav Jain < [EMAIL PROTECTED]>
> > > > > > > > > > > > > > > wrote:
> > > > > > > > > > > > > > > > if u know the range of values stored in array
> > > > > > > > > > > > > > > > then
> > > > > > > > > > > > > > > > let me assume values 1 to k then u can
> > > > > > > > > > > > > > > > calculate sum of
>
> ...
>
> read more >>
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