Array duplication in JavaScript

Question

I want to achieve this

var arr = [1,2,3];

var duplicateArr = duplicate(arr, 2) // will be [1,2,3,1,2,3]

var duplicateArr2 = duplicate(arr, 3) // will be [1,2,3,1,2,3,1,2,3]

I have implemented it like this:

function duplicate(arr, n) {
  var len = arr.length;
  if (n == 0 || n == 1) {
    return arr;
  }
  if (n > 1) {
    for (var j = 0; j < n - 1; j++) {
      for (var i = 0; i < len; i++) {
        arr.push(arr[i])
      }
    }
  }
  console.log(arr);
}
duplicate([1, 2, 3], 3);

I know this may not be the efficient way to do it.

Answer 1

The special treatment for n == 0 and n == 1, and the n > 1 condition are unnecessary. The outer loop condition already handles these cases, so this is roughly equivalent:

function duplicate(arr, n) {
  var len = arr.length;
  for (var j = 1; j < n; j++) {
    for (var i = 0; i < len; i++) {
      arr.push(arr[i]);
    }
  }
  return arr;
}

I say "roughly", because I dropped the console.log.

Answer 2

You can avoid the nested for loops by adding the entire array at the same time. (I'm also using Janos' suggestion here)

function duplicate(arr, n) {
  let result = [];
  for(let i=0; i<n; i++) {
    result.push(...arr);
  }
  return(result);
}

We need the original array unmodified, so we can keep adding it, so I'm using a new result array. This has a few benefits. Firstly it's more common to leave the input array unmodified and return a new array, and secondly it returns an empty array when n=0, which I think makes more sense.

This uses the ES6 spread syntax. If you don't want to use ES6 you can replace

result.push(...arr);

with

result = result.concat(arr);

Answer 3

I think, this will reduce one loop

function duplicate(arr, n) {
    var len = arr.length;
    if (n == 0 || n == 1) {
      return arr;
    }
  if (n > 1) {
    for (var j = 0; j < n - 1; j++) {
      arr = arr.concat(arr);   //**changed here**
     }
   }
 console.log(arr);;
}