arr[]={1,2,4,5,67,89,4}
o(n) storing time into hashtable
index cnt
0 0
1 1
2 1
3 0
4 1(if(h[4] ==0) then enter else dont
5 1
... 0
... 0
67 1
89 1
check before entering into the hash table if the value at that index is 0?
O(1)
if it is then the number has ocured once..increment value in h[i]
if not, then number is occuring for second time..
space required is more...to store table..
but space is cheaper than time...
Sumedh
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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