Are they the “same”? CodeWars

问题

HERE IS THE FULL QUESTION DESCRIPTION

Given two arrays a and b write a function comp(a, b) (compSame(a, b) in Clojure) that checks whether the two arrays have the "same" elements, with the same multiplicities. "Same" means, here, that the elements in b are the elements in a squared, regardless of the order.

Examples

Valid arrays

a = [121, 144, 19, 161, 19, 144, 19, 11]  
b = [121, 14641, 20736, 361, 25921, 361, 20736, 361]

comp(a, b) returns true because in b:

1. 121 is the square of 11,
2. 14641 is the square of 121,
3. 20736 the square of 144,
4. 361 the square of 19,
5. 25921 the square of 161, and so on.

It gets obvious if we write b's elements in terms of squares:

Invalid arrays

a = [121, 144, 19, 161, 19, 144, 19, 11] 
b = [11*11, 121*121, 144*144, 19*19, 161*161, 19*19, 144*144, 19*19]

If we change the first number to something else, comp may not return true anymore:

a = [121, 144, 19, 161, 19, 144, 19, 11]  
b = [132, 14641, 20736, 361, 25921, 361, 20736, 361]

comp(a,b) returns false because in b, 132 is not the square of any number of a.

a = [121, 144, 19, 161, 19, 144, 19, 11]  
b = [121, 14641, 20736, 36100, 25921, 361, 20736, 361]

comp(a,b) returns false because in b, 36100 is not the square of any number of a.

Remarks

1. a or b might be [] (all languages).
2. a or b might be nil or null or None (except in Haskell, Elixir, C++, Rust).
3. If a or b are nil (or null or None), the problem doesn't make sense so return false.
4. If a or b are empty the result is evident by itself.

Note for C

1. The two arrays have the same size (> 0) given as parameter in function comp.

MY QUESTION:

Can you come up with a test case where I do not meet the desired specefications??

I am stuck on 1 basic test not being passed (expected result: true but my code returns false)

MY CODE ATTEMPT

function isTrue(el){
  return el === true;
}

function comp(array1, array2){
  if(array1.length === 0 || array2.length === 0){
    return false;
  }
  var arr = array1.map(function(num){return num*num});
  var arr2 = [];
  for(var i = 0; i < arr.length; i++){
    if(array2.includes(arr[i])){
      arr2.push(true);
      var a = array2.indexOf(arr[i]);
     array2.splice(a,1);
    } else{
      arr2.push(false);
    }
  }
  return arr2.includes(false) ? false : true;
}

回答1:

The easiest possible way:

const comp = (a1, a2) => {
  if (!a1 || !a2 || a1.length !== a2.length) return false;
  return a1.map(x => x * x).sort().toString() === a2.sort().toString();
}

来源：https://stackoverflow.com/questions/45048414/are-they-the-same-codewars