Given a non-empty array of integers, every element appears twice except for one. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Example 1

Input: [2,2,1]
Output: 1

Example 2

Input: [4,1,2,1,2]
Output: 4

Solution

A simple approach using extra space is to use a HashMap of the integers as keys and counts as value. Then iterate over the HashMap to find the key with value 1. That will give us a time complexity of O(n) and space complexity of O(n), n being the size of the array.

Another more simple approach is to use XOR (bit manipulation). Given two numbers, a and b, as we know an XOR a is 0 and 0 ^ b is b. So we can iterate over the array and keep XORing the elements with one another, similar ones will cancel out and we will get the unique one.

/**
 * SingleNumber.java
 * Purpose: Find maximum sum contiguous subarray 
 *
 * @author Megha Rastogi
 * @version 1.0 07/13/2020
 */
class SingleNumber {

    /**
     * Find unique number in an array of duplicates 
     *
     * @param Array, nums
     * @return integer, ans
     */
    public int singleNumber(int[] nums) {
       
        int ans = 0;
        
        // looping through the array and using XOR
        // as a ^ a is 0 and 0 ^ b is b
        for(int i = 0; i < nums.length; i++){
            ans ^= nums[i];
        }
        return ans;
    }
}

Source – leetcode

Code – SingleNumber.java