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
>
> >
>
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