CareerCup

for(int i=0; i<n; i++)
{
 if(a[i] % 2 !=0 )
 { 
   printf("%d", a[i]);
   break;
 }
}

@Anonymous: I think you got the question wrong.

The array has numbers which are repeated even no of times expect one.
ex: {1,1,2,2,2,2,3,3,3,7,7,7,7}

The best way to find out the odd repetition is to do a XOR.

read each array element.
Add element into a hashtable and remove it when u encounter it in the array again.
the elements left in hastable after u are done reading the entire array are the odd elements.

just use XOR. numbers that show up an even number of times will cancel. the value of final will reflect the odd element.

int final = 0;
for (int i=0; i<n; i++)
  final ^= a[i];

check a[0] value with remaining values. if equal increment x. similarly check a[1],a[2]........... . if x value remains odd we can find out the odd repeated value

What if there is no odd element? Example, [1,1,2,2,3,3,4,4] - XORing all this together will give you 0, the same answer you would get by XORing an array where 0 is the odd element.

sort the array.
then push the element in stack. If u encounter the same element , then pop it from the stack.Keep repeating it. The elements that are present even no. of times will be finally removed from the stack and only element present odd no. of times be left in stack