If the numbers in the array continuous .. then it is very simple by using
Arithmetic Progression . we can sum up the all the numbers in an array and
subtract Arithmetic Progression sum..we will get...O(n)..
or
in one trace we can find min and max numbers in that array..
bitset <(max-min)> duplicates;
for(int i=0;i<n;i++)
{
if(duplicates[array[i]-min] ==1)
print duplicate...
else duplicate[array[i]-min]=1;
}
On Wed, Aug 31, 2011 at 2:02 AM, Navneet Gupta <navneetn...@gmail.com>wrote:
> You are given an array. One integer is in the array twice and others
> are unique. Find that no. O(n) Solution
>
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> Navneet
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