JS code to return n times an array not working

Question

The following coding problem has me stumped. There are four tests, which begin with "// ex." They are followed by my code. My code is only passing the first test. I am unsure why it is failing the next 3. Thanks in advance for any insight.

// 1.4 repeat(n, array)
// Write a function that takes a non-negative integer n and an array and           returns a new
// array that contains the contents of given array repeated n times.
// ex. repeat(0, [1]) -> []
// ex. repeat(10, []) -> []
// ex. repeat(1, [1, 2, 3]) -> [1, 2, 3]
// ex. repeat(3, [1, 2, 3]) -> [1, 2, 3, 1, 2, 3, 1, 2, 3]

toolbox.repeat = function(n, array) {
  var arr = [];
  for(var i = 0; i < n; i++){
    arr.push(array);
  }
  return arr;
}

You can print out the result array to some where and see the different. It's basic problem. — KAngel7
– KAngel7, Commented Oct 14, 2016 at 3:06
You're creating an array of arrays, but the function is supposed to return an array of primitives. — david25272
– david25272, Commented Oct 14, 2016 at 3:14

Community · Accepted Answer · 2017-05-23 12:00:40Z

5

Your problem is this line:

arr.push(array);

What this does is pushes the entire array as a single element, so you get:

[ [], [], ... n times ]

But what you want to do is push the contents of the array, n times.

One way you can solve this is to, instead of pushing, you concat array to arr on each iteration:

arr = arr.concat(array);

Another way would be to call Array.prototype.push and pass arr as the context into which to push and array as the arguments to be pushed:

Array.prototype.push.apply(arr, array);

Or in ES6 you can use the spread operator to do the same:

arr.push(...array)

As self demonstrated, you can also create an inner loop to, for every n, iterate over array and add its elements to arr.

edited May 23, 2017 at 12:00

CommunityBot

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answered Oct 14, 2016 at 3:06

nem035

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2 Comments

Ryan Over a year ago

This is certainly maybe more idiomatic. But I think it hides the implementation which may not be the best for this exercise.

nem035 Over a year ago

yeah, that makes sense. I guess it depends on the exact purpose of the exercise but that's why I referenced your answer as well :)

Rajesh Dixit · Accepted Answer · 2016-10-14 03:08:25Z

3

You should use array.concat instead

var repeat = function(n, array) {
  var arr = []
  for(var i = 0; i < n; i++){
    arr = arr.concat(array);
  }
  return arr;
}

console.log(repeat(3,[1,2,3]))

answered Oct 14, 2016 at 3:08

Rajesh Dixit

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Comments

Ryan · Accepted Answer · 2016-10-14 03:22:06Z

1

It looks like you need to push every value of array onto your new array n times. So you'll have an inner loop of:

var repeat = function(n, array) {
  var arr = [];
  for(var i = 0; i < n; i++) {
    for(var j = 0; j < array.length; j++) {
        arr.push(array[j]);
    }
  }
  console.log(arr);
}

repeat(0, [1])
repeat(10, [])
repeat(1, [1, 2, 3])
repeat(3, [1, 2, 3])

edited Oct 14, 2016 at 3:22

answered Oct 14, 2016 at 3:04

Ryan

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4 Comments

nem035 Over a year ago

Unfortunately this won't work, OP needs to add the entire array to arr n times, not just attempt to copy contents once (assuming n is within bounds)

nem035 Over a year ago

oh, you meant two loops, got ya, my bad.

Ryan Over a year ago

Check my edit, i didn't realize he didn't have the first loop

Alex Over a year ago

Both of these methods were extremely helpful, thank you to everyone

Alongkorn · Accepted Answer · 2016-10-14 05:32:47Z

0

your code is incorrect, please change a little bit to

toolbox.repeat = function(n, array) {

  var arr = [];

  for(var i = 0; i < n; i++) {
    // use concat array instead
    arr = arr.concat(array);

    return arr;
}

// repeat(0, [1]) -> []
// repeat(10, []) -> []
// repeat(1, [1, 2, 3]) -> [1, 2, 3]
// repeat(3, [1, 2, 3]) -> [1, 2, 3, 1, 2, 3, 1, 2, 3]

answered Oct 14, 2016 at 5:32

Alongkorn

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