Seek and destroy (intermediate algo scripting)

JavaScript
#1

Tell us what’s happening:

I was able to eliminate the first instance of the arguments, however i can’t eliminate multiple instances. I think I might have to count the number of instances each arg shows up in the array but I’m not sure how to do this optimally.

Your code so far


function destroyer(arr) {
// Remove all the values

for(let i = 1; i < arguments.length;i++){  
    if(arr.includes(arguments[i])){
      const index = arr.indexOf(arguments[i]);
      if (index > -1) {
        arr.splice(index, 1);
      }
    }    
}
console.log(arr);
}

destroyer([1, 2, 3, 1, 2, 3], 2, 3);

Your browser information:

User Agent is: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_0) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/74.0.3729.131 Safari/537.36.

Challenge: Seek and Destroy

Link to the challenge:
https://www.freecodecamp.org/learn/javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/seek-and-destroy

#2

The problem with your code is that splice mutates (changes) the arr and now i points to the wrong position.
One fix would be to add i-- after the line where you do splice.

Also I’m not sure why there is a note that you have to use arguments. Because one could use rest operator (...) which, imho, is much readable.

function destroyer(arr, ...rest) {
  return arr.filter(n => !rest.includes(n))
}
1 Like
#3

Thank you! After looking at the documentation for the rest operator I don’t know why the note said to use arguments either. Your solution is very readable and optimal.

Edit: In the challenge it does not allow to add an additional parameter to the destroyer function, which is why I’m assuming they advised using arguments.

#4

Adding rest operator to function parameters is equivalent to using arguments.
And my solution passes the tests and is accepted even without using arguments explicitly ¯\_(ツ)_/¯

#5