You are given an integer array, where all numbers except for ONE numbers appear even number of times.

Q: Find out the ONE number which appear odd number of times.

This is very simple using bitwise which is going to use only O(1) space.

**Input :** int a[] = { 4, 2, 6, 3, 4, 5, 6, 5, 2 };

**Output :** Number is : 3

public class BitwiseFindOneOddNumber { public static void main(String[] args) { int a[] = { 4, 2, 6, 3, 4, 5, 6, 5, 2 }; System.out.println(" Number is : " + oddNumberIs(a)); } public static int oddNumberIs(int a[]) { int xor = a[0]; for (int i = 1; i < a.length; i++) { xor ^= a[i]; } return xor; } }