*//this method can be applied only when
the value of elements in array are not very large
int count[];
for(i = 0 to n-1)
{
if(count[arr[i]] == 1)
printf(" %d ", arr[i]);
else
count[arr[i]]++;
}*
..
On Wed, Aug 31, 2011 at 12:38 PM, Yuchen Liao <lycdra...@gmail.com> wrote:
> Hi Akshat,
>
> I think that you are wrong here. For example, I have 1, 1, 2 in the
> array. The output will be 2. But not 1.
>
>
>
> On Wed, Aug 31, 2011 at 1:57 AM, Akshat Sapra <sapraaks...@gmail.com>wrote:
>
>> Solution:
>>
>> arr[n],sum = 0;
>>
>>
>> for ( int i = 0 ; i < n; i++ ) {
>> sum ^= arr[i];
>> }
>>
>> print sum; // required number
>>
>> --
>>
>>
>> Akshat Sapra
>> Under Graduation(B.Tech)
>> IIIT-Allahabad(Amethi Campus)
>> --------------------------------------
>> sapraaks...@gmail.com
>> akshatsapr...@gmail.com
>> rit2009...@iiita.ac.in
>> sapraaks...@facebook.com
>>
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>
>
>
> --
> from Yuchen Liao via Gmail
>
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