The xor approach only works if there are no values which occur only
once. But the problem statement indicates that some numbers occur
once, some occur twice, and one occurs three times. So you will end up
with prod equal to the xor of all of the values which occur once or
three times. Put that in your input array and you'll find that you
don't get the desired output.
I don't know of a solution better than sorting and scanning the array.
Don
On Feb 12, 3:14 pm, prakhar singh <prakharsngh.1...@gmail.com> wrote:
> #include<stdio.h>
>
> int main()
> {
> int a[] = {2,2,3,3,3,1,1,4,4}; // is this correct?
> int prod=a[0];int i;
> for(i=1;i<(int)sizeof(a)/sizeof(a[0]);i++)
> {
> prod ^= a[i];
> }
> printf("%d\n",prod); //outputs 3, algorithm works as Sachin described
> it;
>
> }
>
> On Tue, Feb 12, 2013 at 11:44 PM, Sachin Chitale
> <sachinchital...@gmail.com>wrote:
>
>
>
>
>
>
>
> > use ex-or operation for all array elements..
> > a^a=0
> > a^a^a=a
>
> > On Sun, Feb 10, 2013 at 8:22 AM, Mohanabalan D B
> > <mohanabala...@gmail.com>wrote:
>
> >> can use counting sort
>
> >> On Sun, Jul 15, 2012 at 6:37 PM, santosh thota
> >> <santoshthot...@gmail.com>wrote:
>
> >>> If we can retrieve ith prime efficiently, we can do the following...
> >>> 1.maintain a prod=1, start from 1st element, say a[0]=n find n th prime
> >>> 2.check if (prod% (ith_prime * ith_prime )==0) then return i;
> >>> else prod=prod*ith_prime;
> >>> 3.repeat it till end
>
> >>> On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:
>
> >>>> Given an array of integers where some numbers repeat once, some numbers
> >>>> repeat twice and only one number repeats thrice, how do you find the
> >>>> number
> >>>> that gets repeated 3 times?
>
> >>>> Does this problem have an O(n) time and O(1) space solution?
> >>>> No hashmaps please!
>
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> > --
> > Regards,
> > Sachin Chitale
> > Application Engineer SCJP, SCWCD
> > Contact# : +91 8086284349, 9892159511
> > Oracle Corporation
>
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