w3resource

JavaScript - Find the non-repeated element from an array

JavaScript Bit Manipulation: Exercise-14 with Solution

In an array every element appears twice except for one. Write a JavaScript program to find the non-repeated element from the said array using bit-manipulation.

Test Data:
([1]) -> 1
([1, 2, 3]) -> 0 [All elements are non- repeated]
[1, 2, 8, 3, 1, 2, 3, 8, 6, 6, 7] -> 7

Sample Solution:

JavaScript Code:

const non_repeated_num = (nums) => {
   r = 0
   for(let i = 0; i <= nums.length; i++)
     {
       r = r ^ nums[i]
     }
   return r
 }
nums = [1, 2, 8, 3, 1, 2, 3, 8, 6, 6, 7]
console.log(non_repeated_num(nums))

Sample Output:

7

Flowchart:

Flowchart: JavaScript - Find the non-repeated element from an array.

Live Demo:

See the Pen javascript-bit-manipulation-exercise-14 by w3resource (@w3resource) on CodePen.


* To run the code mouse over on Result panel and click on 'RERUN' button.*

Improve this sample solution and post your code through Disqus

Previous: Parity of a given number.
Next: Maximum, minimum of two integers.

What is the difficulty level of this exercise?

Test your Programming skills with w3resource's quiz.



Follow us on Facebook and Twitter for latest update.

JavaScript: Tips of the Day

Returns the sum of an array, after mapping each element to a value using the provided function

Example:

const tips_sumBy = (arr, fn) =>
  arr.map(typeof fn === 'function' ? fn : val => val[fn]).reduce((acc, val) => acc + val, 0);

console.log(tips_sumBy([{ n: 2 }, { n: 4 }, { n: 6 }, { n: 8 }], o => o.n));
console.log(tips_sumBy([{ n: 1 }, { n: 3 }, { n: 5 }, { n: 7 }], 'n'));

Output:

20
16