Coderslang: Become a Software Engineer

Posted on Jun 23, 2021 • Originally published at learn.coderslang.com

JavaScript Interview Question #49: Add a new array element by index

#beginners #codenewbie #webdev #javascript

Will the length of the JS array change? What’s the output?

All JavaScript arrays have the push function. It’s used to add new elements to the array:

const arr = [ 1, 2 ];

arr.push(3);   // [ 1, 2, 3]
arr.push(500); // [ 1, 2, 3, 500]

You can also use an array index to read a certain element or modify it:

const arr = [ 1, 2 ];

arr[0] = 123;

console.log(arr); // [ 123, 2]

But what if the length of an array equals 4, and we try to "modify" the sixth element?

JavaScript in this case is very liberal and allows us to shoot our own foot. The new element will be added into the array and the length will change.

But there’s a surprise! Take a look:

Same code with additional logging:

const arr = [ 1, 2, 3, 4 ];
arr[5] = 'Hello, world!';

console.log(arr); // [ 1, 2, 3, 4, <1 empty item>, 'Hello, world!' ]
console.log(arr.length); // 6

ANSWER: The length of the array will change, and the number 6 will be displayed on the screen.

Learn Full-Stack JavaScript

Top comments (5)

teamradhq • Jun 24 '21

I'd advise you consider assigning an element directly to an array index to be an anti-pattern and bad practice. As shown by your examples, this produces unexpected results and doesn't meet our requirements.

f we say add a **new array element* by index, if an array has length n then we should expect the result to be an array of length n + 1.

Both of these examples fail to meet that expectation:

let arr;

// n + 2 
arr = [ 1, 2, 3, 4]; // (4)
arr[5] = 5; // (6) [ 1, 2, 3, 4, <1 empty item>, 5 ]

// n + 0
arr = [ 1, 2, 3, 4]; // (4)
arr[3] = 5; // (4) [ 1, 2, 3, 5 ]

Another issue is that might might overwrite an existing element:

const arr = [1, 2, 3];
arr[2] = 4; // [1, 2, 4]

I don't think overwriting elements or inserting empty ones is ever what we would want to do...

In my opinion, a better implementation for this interview question would be to write a function to splice a new element into array or add it to the end of the array:

function insertElementAtIndex(arr, index, element) {
  if (index < 1) {
    return [element, ...arr];
  }

  if (index >= arr.length) {
    return [...arr, element];
  }

    const output = [...arr];
  output.splice(index, 0, element);

  return output;
}

Something like this would give us the expected result of an array with a newly inserted element:

insertElementAtIndex([1, 2], -5, 'insert'); // (3) ["insert", 1, 2]
insertElementAtIndex([1, 2],  0, 'insert'); // (3) ["insert", 1, 2]
insertElementAtIndex([1, 2],  1, 'insert'); // (3) [1, "insert", 2]
insertElementAtIndex([1, 2],  2, 'insert'); // (3) [1, 2, "insert"]
insertElementAtIndex([1, 2],  5, 'insert'); // (3) [1, 2, "insert"]

For bonus points, you could allow insertion from end of array when a negative value is passed for index, and allow inserting multiple elements:

function insertelementsAtIndex(arr, position, ...elements) {
  const isOutOfRange = Math.abs(position) >= arr.length;
  const isNegative = position < 0;

  if (position === 0 || (isNegative && isOutOfRange)) {
    return [...elements, ...arr];
  }

  if (isOutOfRange) {
    return [...arr, ...elements];
  }

  const index = isNegative ? arr.length + position : position;
    const output = [...arr];
  output.splice(index, 0, ...elements);

  return output;
}

If we pass a negative position to insertelementsAtIndex, we insert the element before the nth last element, and if it's out of range, we insert it at the start of the array... so -1 inserts before the last item, -2 before the second last item, etc:

insertelementsAtIndex([1, 2, 3], -5, 'insert'); // (4) ["insert", 1, 2, 3]
insertelementsAtIndex([1, 2, 3], -2, 'insert'); // (4) [1, "insert", 2, 3]
insertelementsAtIndex([1, 2, 3], -1, 'insert'); // (4) [1, 2, "insert", 3]

And if we pass an array of length n, and x new elements to insertelementsAtIndex, we should expect an array of length n + x:

insertelementsAtIndex([1, 2, 3], -5, 'a', 'b'); // (5) ["a", "b", 1, 2, 3]
insertelementsAtIndex([1, 2, 3], -2, 'a', 'b'); // (5) [1, "a", "b", 2, 3]
insertelementsAtIndex([1, 2, 3], -1, 'a', 'b'); // (5) [1, 2, "a", "b", 3]
insertelementsAtIndex([1, 2, 3],  0, 'a', 'b'); // (5) ["a", "b", 1, 2, 3]
insertelementsAtIndex([1, 2, 3],  5, 'a', 'b'); // (5) [1, 2, 3, "a", "b"]

If this was an interview question, I suspect it would be intended to verify that you avoid assigning to element index explicitly and instead provide some implementation similar to either of these two examples.

Coderslang: Become a Software Engineer • Jun 25 '21

Great comment! The original question is not about teaching beginners "how to", but rather about explaining the situation "what if".

And the best way to add a new element to the end of the JS array is by using a push method.

Vaibhav • Jul 6 '21 • Edited

const arr = [1,2,3]
 const elements = "insert"
 const index = 2;
 const output = [...arr];
 console.log( output.splice(index, 0, ...elements))

Hey i was just checking your code out, it works but why splice method for me returns an empty array but works for you.

teamradhq • Jul 7 '21 • Edited

It's because Array.splice deletes and inserts items in array, then returns any items that were deleted. So if we don't delete any items, return value of Array.splice(n, 0, ...items) will always be empty [].

In your example, you just need to log output as well as result of splice call:

const arr = [1,2,3]
const elements = "insert"
const index = 2;
const output = [...arr];

console.log( output.splice(index, 0, ...elements)) // []
console.log(output);
// (9) [1, 2, "i", "n", "s", "e", "r", "t", 3]

And if you're wondering why you see individual characters from elements variable; it's because you're using rest parameter, which will spread iterable contents into function args. In case of string this is the individual characters of its value.

If this isn't what you want, and instead you just want to insert string value, you just need to remove rest operator:

const arr = [1,2,3]
const element = "insert"
const index = 2;
const output = [...arr];

output.splice(index, 0, element)
console.log(output);
// [1, 2, "insert", 3]

This is also the reason why we spread arr into a new output array, before calling splice, because it mutates the array it's called on. When designing our implementation, it's better to avoid mutating input and let the consumer decide if they want to mutate.

For example, consumer "mutates" by assigning output of someFn to previously declared variable arr:

let arr = [1, 2, 3, 4];
arr = someFn(arr); 
console.log(arr);
['some', "mutation", 'has', 'occured'];

I think the fact that splice is mutative leads a lot of people to avoid it. However, it's really useful! Here's some of the things you can do with it...

Remove items from array:

const arr = arr = [1, 2, 3, 4];
const result = arr.splice(1,2);
console.log(arr, result);
// (2) [1, 4] (2) [2, 3]

Add items to array:

const arr = [1, 2, 3, 4];
const result = arr.splice(1, 0, 5, 6);
console.log(arr, result);
// (6) [1, 5, 6, 2, 3, 4] []

Substitute items in array:

arr = [1, 2, 3, 4];
result = arr.splice(1, 2, 5, 6);
console.log(arr, result)
// (4) [1, 5, 6, 4] (2) [2, 3]

So to summarise:

Array.splice mutates its array.
Array.splice returns items that were deleted from its array.
Array.splice syntax is:

Array.splice(start, deleteCount, ...items);

Hope this helps explain it more ;)

Vaibhav • Jul 16 '21

Thank You for long explanantion, everything make sense now. I havent used these methods very often so forgot there nature and what it does and for some reason i havent noticed that it mutates the original array it self in mdn documentation. Stupid me...

DEV Community

JavaScript Interview Question #49: Add a new array element by index

Top comments (5)

Read next

Real-Time Processing of Big Data: Tools and Best Practices

Detailed Comparison of Prettier, Biomejs, and VoidZero

Building React Native In-App Modules bypassing the JS bridge.

Web APIs