square of first array individual elements is equal to second array of individual element irrespective of order

Frequency counter pattern

Following is sample javascript code.

function same(arr1, arr2) {
  if(arr1.length !== arr2.length) {
    return false
  }
  let obj = {arr1: {}, arr2:{}, count:0}
   arr1.forEach((item,index) => {
    obj.arr1[item*item] = item*item
  console.log(arr2[index])
    obj.arr2[arr2[index]] = arr2[index]
    
  })
  console.log(obj.arr1, obj.arr2)
  arr1.forEach((item)=>{
    //console.log(obj.arr1, obj.arr2)
   // console.log(obj.arr1[item], obj.arr2[item])
     console.log(item , obj.arr1[item*item] === obj.arr2[item*item])
    if(obj.arr1[item*item] === obj.arr2[item*item]) {
     obj.count +=1
    }else {
      return false;
    }
  })
  
  // console.log(obj)
  return obj.count >=arr1.length ? "true" :"false";
  
}

console.log(same([1,2,3], [1,4,10]))

input same([1,2,3], [1,4,9])
output true
input same([1,2,3], [1,4])
output false
input same([1,2,3], [1,4,7777])
output false