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 (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...
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 > > > > > > > > numbers stored > > > > > > > > > in > > > > > > > > > > > array in O(n) complexity. > > > > > > > > > > > after that apply formula > > > > > > > > > > > > duplicate value= [k*(k+1)]/2 - sum of numbers stored in > > > > > > > > array > > > > > > > > > > > > it will take O(1) constant time so total complexity > > > > > > > > > > > becomes > > > > > > > > only O(n). > > > > > > > > > > > > it can be one solution to your problem but if the range is > > > > > > > > unknown for > > > > > > > > > > > values then > > > > > > > > > > > is there any solution to come in O(n)??? > > > > > > > > > > > > On 8/16/07, dsha < [EMAIL PROTECTED]> wrote: > > > > > > > > > > > > > Hi there, > > > > > > > > > > > > > I'm interested in the following problem: there is an > > > > > > > > > > > > array > > > > > > > > of integers > > > > > > > > > > > > that contains each element only once except for one > > > > > > > > element that > > > > > > > > > > > > occurs exactly twice. Is there a way to find this > > > > > > > > > > > > element > > > > > > > > faster than > > > > > > > > > > > > O(n*log n) and with constant extra memory? If no, how > > > > > > > > > > > > can > > > > > > > > I prove it? > > > > > > > > > > > > > Thanks in advance for ideas. > > > > > > > > > > > > -- > > > > > > > > > > > Vaibhav Jain > > > > > > > > > > -- > > > > > > > > > Vaibhav Jain > > > > > > > -- > > > > > > Vaibhav Jain > > > > > -- > > > > Vaibhav Jain --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---