how will I find odd elements from two arrays

Question

I have one array

int a[] = {1,2,3};
int b[] = {1,2};

how will I find the odd element in array

Answer 1

Make a loop for each array item. Those for which item & 1 == 1 are odd.

Answer 2

    int[] longArray = { 1, 3, 2 };
 int[] shortArray = { 1, 2 };

 //Check which array is longer, if b longer than a then swap

 boolean found = false;
 int odd = 0;

 for (int i : longArray) {
  for (int j : shortArray) {
   if (i == j)
    found = true;
  }
  if (!found)
   odd = i;
  found = false;
 }
 System.out.println(odd);

Answer 3

If you are talking about Set Comparisons, have a look at java-is-there-an-easy-quick-way-to-and-or-or-xor-together-sets

This SO question was talking in terms of the Set interface, but the List interface also inherits the relevant methods from Collection, so copying your arrays to ArrayList objects makes it easy.

If you want to keep it purely at the native array level, you might want to do something like:

 public int[] findOddElement(int[] fromArray, int[] secondArray){
  int[] result = new int[fromArray.length];
  int resPointer = 0;
  for (int i = 0;i < fromArray.length;i++){
   boolean notFound = true;
   for (int j = 0; j < secondArray.length; j++) {
    if (fromArray[i] == secondArray[j]) {
     notFound = false;
     break;
    }
   }
   if (notFound){
    result[resPointer] = fromArray[i];
    resPointer++;
   }
  }
  if (resPointer > 0 && resPointer < fromArray.length ) {
   int[] newResult = new int[resPointer];
   for (int i = 0;i < resPointer; i++) {
    newResult[i] = result[i];

   }
   return newResult;
  }
  return result;
 }

Answer 4

    int[] longArray = { 1, 3, 2 };
    int[] shortArray = { 1, 2 };

    //Check which array is longer, if b longer than a then swap

 for (int x:longArray){
        set.add(x);
 }
 for (int x:shortArray){
     if (set.contains(x))
        set.remove(x);
 }
 //odd numbers
 for (Object i:set.toArray())
  System.out.println(i+",");

I put two answers for the voting, the previous is nested loops and might be slow for very long arrays. In the next solution some might dont prefer the TreeSet.