Intermediate Algorithm Scripting - Seek and Destroy

JavaScript
1

Can someone tell me why this code does not work

function destroyer(arr) {
  let arr2= Array.from(arguments).slice(1);
  let arr3=[];
  for (let i=0; i < arr; i++)
  for (let j=0; j < arr2; j++) {
  if (arr[i]!==arr2[j]) {
arr3.push (arr[i])
  } 
  }
 return arr3
};

console.log(destroyer([1, 2, 3, 1, 2, 3], 2, 3));

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Challenge: Intermediate Algorithm Scripting - Seek and Destroy

Link to the challenge:

2

Indenting your code:

function destroyer(arr) {
  let arr2= Array.from(arguments).slice(1);
  let arr3=[];
  for (let i=0; i < arr; i++) {
    for (let j=0; j < arr2; j++) {
      if (arr[i] !== arr2[j]) {
        arr3.push(arr[i]);
      } 
    }
  }
  return arr3;
}

I would double check your logic. Every single time arr[i] is not equal to one of the elements of arr2 you add it to the arr3? Have you seen how much stuff you’re putting into arr3?

3

The console just says ‘’ every time. Is it okay if you just spell it out what I am doing wrong because I am still confused to be honest

4

Also need to fix this - arr2 isn’t a number

I’m trying to do that. I’m still cannot give you the solution code.

5

Can you give me any more information?

6

What have you done about this? What sort of thing is arr2?

7

yes I have changed this to arr2.length

8

and here too

9

Yep have done that one now

10

Ok. So that brings us around to

What’s coming into arr3 now?

11

1,1,2,3,1,1,2,3. It has doubled the 1s, joined arr 1 and arr2 together, and taken a two out… Not sure why

12

This says to add arr[i] to the array arr3 every single time it doesn’t match one of the elements in arr2. I think you want instead to add arr[i] to arr3 only if it doesn’t match every single element in arr2.

13

Given what you said, I changed it to this and it worked (apologies if formatting is bad)-

function destroyer(arr) {
  let arr2= Array.from(arguments).slice(1);
  let arr3=[];
  for (let i=0; i < arr.length; i++) {
    for (let j=0; j < arr2.length; j++) {
      if (arr[i] === arr2[j]) {
        arr3.push(arr[i]);
      } 
    }
  }
return arr.filter(val => !arr3.includes (val))
};
14

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