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XOR all the numbers together you will get the odd one out. O(n)

Thanks almost missed that approach ,I used hashtable . But never the less i could clear the round

int a[]={1,2,3,4,5,6,7,8,9,1,2,3,5,6,7,8,9};
int size=sizeof(a)/sizeof(int);
int sum=0,p;
for(int i=0;i<size;i++)
printf("%d ",a[i]);
for(int i=0;i<size;i++)
{
sum=sum^a[i];
}
printf("\nnun duplicated number is %d",sum);

Yeah, I would use a hash table too. My values would be booleans which I would simply toggle. Whichever value is true after parsing the array, the key would be that number.

need to check just half of the array.

garimaa.blogspot.in/2012/04/program-12th-in-c_11.html