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);
  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 */
}

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.

1 Like

Used BubbleSort to clear this challenge.
Very doable after the lessons on BubbleSort.

Validating that x is in the array provided the final two checkmarks.
Cheerio!

function getX(x, nums) { //step 1 : sort the list const sort = (array) =>{ let swapping = true; while(swapping){ swapping = false; for(let i = 0; i < array.length -1;i++){ if(array[i] > array[i+1]){ swap(array, i, i+1) swapping = true; } } } return array; } const swap = (array, indexOne, indexTwo) => { let temp = array[indexOne]; array[indexOne] = array[indexTwo]; array[indexTwo] = temp; } //call the sort function to sort the list; sort(nums); if (x > nums.length || x < 1){ return 0; } return nums[x-1]; } console.log(getX(2, [5, 10, -3, -3, 7, 9])); // Leave this so we can test your code: module.exports = getX;
1 Like

Your solution with BubbleSort is radically faster than other
Test: x=500, array with 1000 random integer from -50 to 50 (warming up, 1000 calls of 20 passes)

  1. 0.022 Array.sort
  2. 1.536 janbazant1107978602’s solution with generator
  3. 0.002 Your solution. Wow!
2 Likes
function getX(x, nums) {
  if (!x || x > nums.length) 
    return ''
  const arr = nums.sort((a, b) => a - b)
  return arr[x-1]
}

module.exports = getX;

My idea was not to sort all list because we need only x minimums from the nums array to get the correct number from the list, so we will sort all list only in the worst case when x=nums.length. In the worst case, we got time complexity O(N^2) same as bubble sort, but in other cases, we got an advantage because we don’t need to sort all list.
The outer loop will run x times to find x minimums, each time you find a minimum you push it to the sorted array, record the index of this minimum and remove this minimum from the nums length decrease n-1, and all repeats, we search next minimum until sorted array length become x. I hope my logic didn’t go wrong and there is still some advantage. :slightly_smiling_face:

function getX(x, nums) { // Write your code here const sortedArray = []; let min = nums[0]; let index = 0; if (x > 0 && x <= nums.length) { while (sortedArray.length < x) { for (let i = 0; i < nums.length; i++) { if (min > nums[i + 1]) { min = nums[i + 1]; index = i + 1; } } sortedArray.push(min); nums.splice(index, 1); index = 0; min = nums[0]; } return sortedArray[x - 1]; } else { return 0; } } console.log(getX(2, [5, 10, -3, -3, 7, 9])); // Leave this so we can test your code: module.exports = getX;
function getX(x, nums) {
//if x is NOT in the range of the amount of numbers available in the array, return 0
if(x < 1 || x > nums.length){
   return 0;
 }
 //if it is, sort the array in order then return the number at X location in the array
 else{
   nums.sort((a,b)=>a-b);
 }
return nums[x-1]
}

module.exports = getX;

Here is my code for this chalenge, It took me sometime to understand that in the test, there were situations where x was outside the range of the array.length.
Therefore i decided to first test if x was valid and only sort the numbers once it passed the first test

-here is a more concise alternative, if we don’t mind sorting the array before checking the validity of x

function getX(x, nums) {
//sort nums in ascending order
nums.sort((a,b)=>a-b);
//return 0 if x not in the range of the array, else return number at x-1 position
return (x < 1 || x > nums.length)? 0: nums[x-1]
}
//test code
module.exports = getX;

Used bubble sort.
function getX(x, nums) {
// Write your code here
if (x > 0 && x <= nums.length && nums.length > 0) {
let swapping = true;
while (swapping) {
swapping = false;
for (let i = 0; i < nums.length -1; i++) {
if (nums[i] > nums[i+1]) {
swap(nums, i, i+1);
swapping = true;
}
}
}
console.log(nums);
return nums[x-1];
}
return 0;

}
function swap (arr, index, nextIndex) {
const temp = arr[nextIndex];
arr[nextIndex] = arr[index];
arr[index] = temp;
}

console.log(getX(2, [5, 10, -3, -3, 7, 9]));

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

Hello world This is my soluction of the excersise

function getX(x, nums) {
  nums.sort((a,b) => a - b)
  const result = nums[x - 1]
  return result ? result : 0;
}

function getX(x, nums) { nums.sort((a,b) => a - b) const result = nums[x - 1] return result ? result : 0; } console.log(getX(4, [5, 10, -3 , -3, 7, 9])); console.log(getX(2, [5, 10, -3, -3, 7, 9])); // Leave this so we can test your code: module.exports = getX;

Happy Coding!!

function getX(x, nums) {
if(nums.length===0 || x < 1 || x > nums.length){
return 0
}

let swapping=true;

while (swapping){

swapping=false;
for(let i=0; i < nums.length-1; i++){
  if(nums[i] > nums[i+1]){
    swap(nums, i, i+1);
    swapping=true;
  }
}

}

return nums[x-1];

}

const swap = (arr, index1, index2) =>{
let temp= arr[index2];
arr[index2]=arr[index1];
arr[index1]=temp;
}; // Write your code here

console.log(getX(1, [5, 10, -3, -3, 7, 9]));
console.log(getX(4, [5, 10, -3 , -3, 7, 9]));
// Leave this so we can test your code:
module.exports = getX;