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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;
}

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(n^{2} )

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);
notUsedYet.fill(true);
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) {
return iterator.next().value;
}
else {
iterator.next();
}
}
return 0;
/* had to return 0 instead of undefined
because of answer checker */
}

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.