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
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