Can I solve this in just a single "for" loop. Or is it impossible?

Question

https://code.sololearn.com/ceFVnFidR1ic/?ref=app
I have 2 arrays with same length. Each squared number of first array should have corresponding number in second array and vise versa. i.e.
let arr1 = [2,4];
let arr2 = [4,16] // or [2*2, 4*4];
So I treat them equal.
Elements in arrays can be in random order as in code.
I described my problem in comments of code.

Answer 1

Artur Here's my solution in case you get stuck.

https://code.sololearn.com/ctgeIWy0gP4l/?ref=app

Answer 2

Artur Assuming recursion was also not allowed and efficiency wasn't important, then I suppose you could accomplish this within a single infinite looping construct where the break condition occurred within the loop block.

This single loop would need separate variables to store the current index (i and j) for each respective array with reset logic to be applied to the 2nd array after reaching the array length. The first array index would likely only increment when the 2nd array index was equal to the length of the 2nd array.

You could even attempt to count the items in the array this way.

Consider this super simple code:

let a = [2, 4]
let b = [4, 16]
let i = 0
let j = 0
let i_size = null
let j_size = null

for( ; ; ) { // same as while(true)

//Count length
if( null == i_size ) {
if("undefined" == typeof a[i] ) {
i_size = i - 1//set length
i = 0 //reset i
} else {
i++ //increment i
}
contiue
//Skip everything else below
//until i_size has a value.
}
}

Answer 3

You can iterate <array2>, and find a square root of each number from <array2> inside <array1> using array `indexOf` method. When the `indexOf` method returns -1, you'll know that number from <array2> is no square of either number in <array1>.

for(let n of array2)
{
let needle = Math.sqrt(n);
let pos = array1.indexOf(needle);
console.log(`square root of ${n} -> ${needle} found at ${pos}`);
}

Answer 4

Good idea! But.. there is still array method indexOf(). What I wanted is to do that without any array methods (I know it sounds stupid tho 🤣). With array methods there are a few other ways to solve this easily, i.e.:
array2.every(i => array1.some(j => j * j == i));
Or we could filter() same numbers in both arrays and problem departs.

Answer 5

Okay, I thought you only don't want to use every() or some().

It will be pretty hard to do with a single loop as your title described though 😁

Answer 6

Agree. Might be even impossible to do that w/o methods not just hard. I was just wondered if it's possible.

Answer 7

To add more complexity, you can add "avoid use of length property" to the requirements 😂

Answer 8

That would be the mightiest kidding in the world

Answer 9

I just got inspired by a question in Q&A recently, no hard feelings man 😊

Answer 10

This will keep growing, but hopefully you get the idea.

Answer 11

Wow what an amazing idea. I'll definitely try to implement it. Thank you David Carroll !!

Answer 12

And please don't suggest me to use any of methods like every() or some() etc. I really want to understand how to do this with just "for" loops.