 # Intermediate Algorithm Scripting: Sorted Union- help me please

link to the challenge : https://www.freecodecamp.org/learn/javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sorted-union
why my code is returning an empty array `arr1 = [] ? `
Here is my code :

``````function uniteUnique(arr) {
let arr1 = [];
for(let i = 0; i < arr.length; i++){
for(let j = 0; j < arr[i].length; j++){
if(arr1.indexOf(arr[i][j]) === -1){
arr1.push(arr[i][j]);
}
}
}
return arr1;
}

console.log(uniteUnique([1, 3, 2], [5, 2, 1, 4], [2, 1]));
``````

-This looks like a challenge, can you link the challenge here.
Next Time if you need a help from a challenge, you can use the `Get Help` button on the sidebar.

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oh, thanks man i’ll make sure to do that next time . Here’s the link to the challenge https://www.freecodecamp.org/learn/javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sorted-union

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It outputs 0, because over here:

``````for(let i = 0; i < arr.length; i++){
console.log(arr[i])
for(let j = 0; j < arr[i].length; j++){
console.log(arr[j])
if(arr1.indexOf(arr[i][j]) === -1){
arr1.push(arr[i][j]);

}
}
}

uniteUnique([1, 3, 2], [5, 2, 1, 4], [2, 1]);
``````

You are saying that if `i < arr.length` which is 3 from the `console.log`, because the first argument only have 3 items. Then you are testing if `j < arr[i].length` the length is 1 because `arr` is `1` and it is only a single number without length.

• The problem you are having is that you don’t have anyway to take all of the arguments, because you never know how many arguments is going to be passed through the parameter. You can solve this using the `[...arguments]`. This will grab all the argument from the parameter. Then you can loop through all of them and try to solve the challenge

oh. yeah how did i not notice that mistake ! thanks man i appreciate the help

1 Like