CodeWars Javascript 挑战：它们是否“相同”？

Question

The Kata: link

My solution:

function comp(array1, array2) {
    let result;
    if (Array.isArray(array1) && Array.isArray(array2) && array1.length && array2.length) {
        result = true;
        const squares = array2.map(e => Math.sqrt(e));
        squares.forEach((e) => {
            if (array1.includes(e)) return;
            result = false;
        });
    } else {
        result = false
    }
    return result;
}

I don't want another solution. I want to figure out why mine doesn't pass in all of the tests. (Fails on two tests but I can't see which)

I suspect the test expects true if both arrays are [] . But the Kata's description says otherwise:

If a or b are nil (or null or None), the problem doesn't make sense so return false.

Help would be appreciated.

---

Working Solution Based off of the answers:

function comp(array1, array2) {
    let result;
    if (Array.isArray(array1) && Array.isArray(array2)) {
        result = true;
        const sortedArray1 = array1.sort((a, b) => a - b);
        const squares = array2.map(e => Math.sqrt(e)).sort((a, b) => a - b);
        squares.forEach((e, i) => {
            if (sortedArray1[i] === e) return;
            result = false;
        });
    } else {
        result = false
    }
    return result;
}

Answer 1

Okey, some cheating workarounds but your code fails here: That's why after removing two .length you pass one more test.

Next one is After .sqrt array2 you get 2, 3, 3 that's why you return true, but it's false. That answer will be enough to help you solve this kata.

Btw.

I suspect the test expects true if both arrays are []. But the Kata's description says otherwise:

Kata's description:

a or b might be [] (all languages except R, Shell). : - D

PS I know for some people checking arguments maybe treated as cheating, I just do it for educational purposes. Black boxes are not always enough to figure out what's wrong.

Answer 2

if there is more then one number the same, but different count here and there, it will return true in your code, since it's includes in both arrays. you should sort it and iterate by index for compare, in order to make sure each item in the array used only once.

Here is what works for me:

function comp(array1, array2){
  if(!array1 || !array2) return false;
  array1 = array1.map(t => t**2).sort((a,b)=>a-b);
  array2 = array2.sort((a,b)=>a-b);

  for(let i=0;i<array1.length;i++){if(array1[i] !== array2[i])return false}
  return true;
}

sure youcan understand the idea and just make some twick in your existing code