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
> >
>
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