CareerCup

XOR all the elements of array.

public static void main(String[] args) {
		// TODO Auto-generated method stub
		int[] a = {5,5,5,2,2,7,8,7,8,12,10,10,12};
		int b = a[0];
		for(int i = 1; i < a.length ; i++){
			b = b^a[i];
		}
		System.out.println("--> " + b);

	}

Should we assume every time that the array will contain only positive elements and only one of them will be repeated odd number of times?

Not one of the best.
Suggestions appreciated.

int i=0; sum=0; count=0;
while( a[i] == a[i+1] && (i+1) < n )
{
  sum = sum + a[i]; 
  i++; count++;
}
sum = sum + a[i]; count ++;

if( sum % 2 != 0 )
{
  cout<< a[i]; break;
}
else if ( sum % 2 ==0 && count % 2 != 0 )
{
  cout<< a[i]; break;
}
else
{
  i++; count = 0; sum = 0;
}

@Apurva : XOR the value +ve or -ve 1 1 = 0 0 0 = 1

int arr2[] = {-2,-2,10,10,20,20,20};
int element = arr2[0];
for(int i=1;i<arr2.length;i++) {
element = element ^ arr2[i];
}
System.out.println(element);

Works fine and gives me output 20

How about XOR?

#include <stdio.h>
int
main()
{
  unsigned int a[] = {5, 5, 5, 2, 2, 7, 8, 7, 8, 12, 10, 10, 12};
  unsigned int i = 0, x = 0;

  for(i = 0; i < sizeof(a) / sizeof(int); i++) {
    x ^= a[i];
  }

  printf("%d\n", x);
}

Output:
5

http : // www . testskillshome . com/?p=2276

why does this happen? I mean what's the logic?

Does anyone know if its possible to do a hashmap type something for integers?