Finding the array element that appears once

A few days ago, during a job interview, I was asked to live code three exercises in JavaScript:

  1. Palindrome check
  2. Find the element that appears once
  3. Fizz Buzz

Exercises 1 and 3 went pretty well. I was able to implement them correctly, even though, for Exercise 1, my initial answer didn't cover all the edge cases (forgot about space characters and upper/lower case).

Exercise 2 was a bit trickier for me.

Given an array of integers, all the elements appear twice, except one element that appears only once. Write an algorithm to find that element.

The most obvious, brute force solution, consists of two nested loops, where each element in the array is compared with all the other elements, in order to find which element is not repeated. Time complexity of this solution is O(n^2).

I wanted to avoid an answer with such time complexity - also didn't want to write two nested for loops. So, my first instinct was to look for a higher order function, such as, Array.prototype.find() or Array.prototype.filter(), to which I could pass a function that counted the number of occurrences of an element in the array.

Since I was allowed to check the web for syntax and built-in methods, I started searching for such a method.

I realized there's no built-in method that does this in JavaScript and wasn't finding a simple and elegant solution.

As I decided to implement a working solution, and worry about optimization later, the interviewer rightfully commented about the time complexity of the algorithm I was writing.

In fact, this approach also has time complexity O(n^2).

The interviewer proceeded to suggest creating an object and using the array elements as its properties, to which the corresponding values would keep a counter on the number of occurrences.

I immediately remembered doing something similar before, so I knew Array.prototype.reduce() was the way to go.

As I started to implement, I took a bit too much time to figure up the function I needed to pass to reduce() and I didn't get the chance to finish.

So, here I am to complete this challenge and improve my understanding on how reduce() operates ๐Ÿ˜‰.

This solution has time complexity O(n) and consists of using a hash map to store the count of each element. Then, it iterates through the keys and returns the key which has count 1.

function findElementOnce(arr) {
  let hash = arr.reduce((acc, value) => {
    acc.hasOwnProperty(value) ? acc[value]++ : acc[value] = 1;
    return acc;
  }, {});

  for (const key in hash) {
    if (hash[key] === 1) {
      return key;

Important notes about this solution, which became now, clearer to me:

During my research for this article, I've discovered that this challenge can also be solved efficiently, with time complexity O(n), using the operator XOR.

In fact, it can even be considered more efficient, since in terms of space complexity it has O(1), whilst the hashing solution has O(n).

function findElementOnce(arr) {
  if(arr.length == 0) return;
  let xor = arr[0];
  for (let i = 1; i < arr.length ; i++) {
    xor ^= arr[i];
  return xor;

We know that A XOR A = 0. If we XOR all the elements in the array, the elements which are repeated twice will become 0, while the element which is appearing only once will remain.

There's also a more complex version of this problem, where every element appears three times instead of twice, but I'll leave it for you to dive deeper ๐ŸŒŠ.