Problem on a codewars exercise in JS, Prefill an Array; beginner level

Question

Here is the page link of the exercise: https://www.codewars.com/kata/prefill-an-array/train/javascript

Create a function prefill that returns an array of n elements all having the same value v. See if you can do this without using a loop.

You have to validate input:

v can be anything (primitive or otherwise) if v is omitted, fill the array with undefined if n is 0, return an empty array if n is anything other than an integer or integer-formatted string (e.g. '123') that is >=0, throw a TypeError When throwing a TypeError, the message should be "n is invalid", where you replace n for the actual value passed to the function.

my code:

function prefill(n, v) {

  if(n == 0) return [];

  if(!Number.isInteger(n) || n < 0 ){
    let TypeError = new Error;
    TypeError.name = "TypeError";
    TypeError.message = n + " is invalid";
    throw TypeError
  };

  return new Array(n).fill(v); 
}

It passes all the test except this one: should throw an error with n as boolean Test Passed: Value == 'TypeError' Test Passed: Value == 'true is invalid' Test Passed prefill did not throw an error with n as false

Can anybody explain me why? I even tried inputting directly false and it did not work :/

Maheer Ali · Accepted Answer · 2020-01-08 19:02:09Z

1

You are almost there. Just a simple problem in the first if statement. You need to check strictly. Then you also need to check for '0' because the description of problems allows an integer-formatted string

if(n === 0 || n === '0') return [];

The problem n is false then n == 0 evaluates to true so an empty array is returned. But you need to throw error in case of false.

answered Jan 8, 2020 at 19:02

Maheer Ali

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2 Comments

baloo Over a year ago

Thanks for the explanation, I m still new to coding so I don't see all the subtleties yet ^^'

hc_dev Over a year ago

Although not explicitly stated as problem, the solution does not pass one test in link challenge: Test.assertSimilar(prefill('1',1), [1]); and hence does not fulfill the required anything other than an integer or integer-formatted string (e.g. '123') that is >=0.

Alberto L. Bonfiglio · Accepted Answer · 2020-01-08 19:39:31Z

1

The issue seems to be that Number.isInteger(n) expects a number not a string. If you use parseInt(n) all works.

Also the difference between == and === is that: == converts the variable values to the same type before performing comparison. This is called type coercion. === does not do any type conversion (coercion) and returns true only if both values and types are identical for the two variables being compared.

In general it's best to use === but in your case, since you are comparing integers and strings to 0 you could use if(n == 0) return [];

function prefill(n, v) {
  if(n === 0 || n === '0') return [];

  if(!parseInt(n) || n < 0 ){

    let typeError = new TypeError();
    typeError.name = "TypeError";
    typeError.message =  n + " is invalid";
    throw typeError
  };

  return new Array(n).fill(v); 
}

Test Passed: Value == '[1, 1, 1]'
Test Passed: Value == '[\'abc\', \'abc\']'
Test Passed: Value == '[1]'
Test Passed: Value == '[[\'2d\', \'2d\'], [\'2d\', \'2d\'], [\'2d\', \'2d\']]'
Test Passed: Value == 'TypeError'
Test Passed: Value == 'xyz is invalid'
Test Passed
You have passed all of the tests! :)

edited Jan 8, 2020 at 19:39

answered Jan 8, 2020 at 19:33

Alberto L. Bonfiglio

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1 Comment

hc_dev Over a year ago

Thoroughly solved! Your solution also passes the failing test: Test.assertSimilar(prefill('1',1), [1]);, although not explicitly as problem by the OP.

Gabriel GUERIN · Accepted Answer · 2021-04-02 10:15:06Z

0

The solution above did not pass the "should throw errors with n as a float" test. Instead of using !parseInt(n), I used !Number.isInteger(n) found in this solution.

Here is my full answer:

function prefill(n, v) {
    // Return empty array if 0
    if (n === 0 || n === '0') return [];

    // Throw error if below 0 or not an integer 
    if (n < 0 || !Number.isInteger(n)) {
        let typeError = new TypeError();
        typeError.name = "TypeError";
        typeError.message = n + " is invalid";
        throw typeError
    };

    return new Array(n).fill(v);
}

answered Apr 2, 2021 at 10:15

Gabriel GUERIN

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