Presented here is a solution for the given problem with a time complexity of O(n):

const moveZeroes = (nums) => {
  let count = 0;

  for (let i = 0; i < nums.length; i++) {
    if (nums[i] !== 0) {
      nums[count++] = nums[i];
    }
  }

  for (let i = count; i < nums.length; i++) {
    nums[i] = 0;
  }

  return nums;
};

Like many others have pointed out, if the instructions specify not to return anything, then refrain from including a return statement in your code.

However, there seems to be an issue with the implementation you provided. Consider using the test case [0,0,1]. Starting at the first element will shift it to the back of the array, resulting in [0,1,0]. Consequently, the loop index now focuses on the second element (1) while overlooking the second occurrence of 0.

function rearrangeZeros(n) {
  // initialize an array to store zeros
  let zeros = [];
  
  // filter out zeros from the input array and add them to the zeros array
  n = n.filter(item => (item !== 0 ? true : zeros.push(item) && false));
  
  // concatenate the zeros array to the end of the filtered array
  return n.concat(zeros);
}

In situations where a return statement is required but no value can be returned, simply input

return;

Diving into the Depths of Javascript

Many believe that null represents nothing in Javascript. However, technically speaking, it is not truly nothing. If you were to type typeof null, the result would be "object", indicating that it is something rather than nothing. On the other hand, if you execute a function that does not return anything:

(function(){})();

You will observe undefined displayed in the console. Therefore, whether you use return undefined, simply return, or even omit the return statement altogether, all are considered acceptable.

A Fresh Perspective on Sorting

In considering alternative solutions for sorting, let's take a practical approach and introduce a different viewpoint. Leveraging Javascript allows us to utilize the sort function. Nevertheless, custom callback functions must be created for comparing elements:

array.sort(function(a, b) { ... comparison here ... });

For your specific scenario, let's establish our criteria: We assert that

a is deemed less than b only if a equals zero while b does not equal zero.

Therefore, our callback should look like this:

function(a, b) {
    if(a==0 && b!=0)
        return 1;
    else if(b==0 && a!=0)
        return -1;
    else 
        return 0;
}

Utilized in this manner:

var array = [2,6,0,666,-6,142,0,1,-10];
array.sort(function(a, b) {
        if(a==0 && b!=0)
            return 1;
        else if(b==0 && a!=0)
            return -1;
        else 
            return 0;
    });

This approach is typically employed when developing programs as it enhances comprehension (once accustomed to it) and often results in more concise code.

Implement this method using ES6 functionalities

const removeZeros = nums => {
    const filteredArray = nums.filter(element => element !== 0); // Filter out all zero elements
    const countOfZero = nums.length - filteredArray.length; // Calculate the number of zeros removed
    
    const zeroArray = Array(countOfZero).fill(0); // Create an array filled with zeros equal to count of zero elements
    const newArray = [...filteredArray, ...zeroArray]; // Merge filtered non-zero elements with zero elements array

    return nums.splice(0, nums.length, ...newArray); // Update original array by replacing with new array using spread operator
}

Another method that can be applied involves utilizing two pointers

function rearrange(arr) {
  let leftPointer = 0;
  let rightPointer = arr.length - 1;

  while (rightPointer > leftPointer) {
    if (arr[leftPointer] === 0) {
      // swapping the elements --->
      let lastElement = arr[rightPointer];
      arr[rightPointer] = arr[leftPointer];
      arr[leftPointer] = lastElement;
      rightPointer--;
    } else {
      leftPointer++;
    }
  }

  return;
}

let arrayToModify = [1, 2, 0, 7, 9, 0, 3, 0]
rearrange(arrayToModify)
console.log(arrayToModify);

To populate each slot, you can loop through and assign a value from another index that advances faster by skipping zeroes. When the second index goes out of range, you can make use of the `??` operator:

function moveZeroes(nums) {
    for (let i = 0, j = 0; i < nums.length; i++) {
        while (nums[j] === 0) j++; // skip zeroes
        nums[i] = nums[j++] ?? 0;
    }
}

  const filterZeros = (arr) => {
    const nonZeroArray = [];
    const zeroArray = [];
    for(let i=0; i < arr.length; i++){
        if(arr[i] !== 0){
            nonZeroArray.push(arr[i]);
        }else{
            zeroArray.push(arr[i]);
        }
    }
    return nonZeroArray.concat(zeroArray);
}

const result = filterZeros([2, 6, 0, 9, 1, 0]);
console.log(result);

Check out this nifty function:

const reorderZeros = arr => {
    let tempArr = [...arr];
    let zeroCount = 0;
    for(let i=0;i<arr.length;i++){
        if(arr[i] === 0 ){
            arr.push(arr.splice(i, 1)[0]);
            i--;
            zeroCount++;
        }
        if(i + zeroCount + 1 === arr.length) break;
    }
    return arr;
}

reorderZeros([1,0,2,3,0,0,0,4]) // [1,2,3,4,0,0,0,0]

Whenever a zero is found in the array, it gets moved to the end, causing all other elements to shift left and necessitating a `i--`. The loop exits when either all zeros are moved to the right side or we reach the end of non-zero elements. This clever method ensures the correct reordering of zeros in the array.

The provided technique is not effective when dealing with consecutive zeros, as illustrated below:

nums = [0,0,1]

• At index=0 output ==> [0,1,0]
• At index=1 output ==> [0,1,0] This demonstrates how focusing on index 1, which is non-zero, can lead to missing the zero at index 0 and resulting in incorrect outcomes.

To address this issue, I propose the following revised approach:

   var adjustZeroes = function(nums) {
    let length = nums.length
    if(length<=1)return nums
   let zeroCounter=0;
   for(let j=0;j<length;j++){
       if(nums[j]===0)zeroCounter++
       else if(nums[j]!=0&&zeroCounter!=0){
           nums[j-zeroCounter]=nums[j]
           nums[j]= 0
       }
   } 
};