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
> > >
> >
>
> >
>
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