JavaScript Challenge - Find Xth Number In Order

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A post was merged into an existing topic: Python Challenge - Find Xth Number In Order

It would only accept my version when I did:
when there is no Xth number, return 0 (instead of undefined).

Minimal version using .sort()
although I had to provide a comparison function to it for it to work properly.

const compare = (a, b) => {
  if (a < b) { return -1; }
  else if (a > b) { return 1; }
  else { return 0; }

const getX = (x, nums) => { 
  const result = Array.from(nums).sort(compare)[x - 1];
  return result ? result : 0;
1 Like

Here’s another version where I made an iterator (using a generator function) and did a simple sorting algorithm in it (not in-place).
The algorithm I used is: find min, insert it in new array, find min of remaining stuff, insert it, etc. 0(n2 )

my code (long)

I split the tasks into 3 functions, instead of writing more loops inside loops.

function min(nums, bools, changeMinToFalse) {
  // finds minimum only for positions where bools has true
  let minimum = undefined;
  let indexOfMin = undefined;
  const length = nums.length;
  for (let i = 0; i < length; i++) {
    if ((bools == undefined) || bools[i]) {
      if ((minimum == undefined) || (nums[i] < minimum)) {
        minimum = nums[i];
        indexOfMin = i;
  if (changeMinToFalse && !(indexOfMin == undefined)) {
    bools[indexOfMin] = false;
  return minimum;

function* inOrder(arr) {
  // generator function to iterate from least to greatest
  const length = arr.length;
  const notUsedYet = new Array(length);
  for (let i = 0; i < length; i++) {
    yield min(arr, notUsedYet, true);

function getX(x, nums) {
  const length = nums.length;
  if (x < 0) { x = length + x; }
  const iterator = inOrder(nums);
  for (let i = 1; i <= length; i++) {
    if (i == x) {
    else {;
  return 0;
  /* had to return 0 instead of undefined 
     because of answer checker */

Here is my example code. I needed to add the sort function check manually for it to work. !

 function getX(x, nums) {

// Write your code here

   const sorty = (a, b) => {
    return a - b; 


    let arr = nums.sort(sorty);
     let answer = 0; 

     arr.forEach((value, iterator) => {
     if(iterator === x - 1){
       answer = value; 

  return answer; 

  console.log(getX(4, [5, 10, -3, -3, 7, 9]));

  // Leave this so we can test your code:
  module.exports = getX;

const getX = (x, nums) => x < 1 || x > nums.length ? 0 : nums.sort((a,b) => a-b)[x-1]

1 Like
function getX(x, nums) {

 nums.sort((a, b) => a - b);
  if(x < 1 || x > nums.length){
   return 0;

return nums[x - 1];

module.exports = getX;

Thank you! I returned “Error” instead of 0 for broken inputs and failed testcases and I couldn’t find out why. The instructions missing some points. As someone who learned C, returning 0 when the function is broken is very painful.