Discussion on: Simplest way to compare two numbers array in JS

[Kavindu Santhusa]

Correct Algorithm

Here is the Correct Algorithm

const array1 = [1, 2, 3, 4, 5];
const array2 = [1, 2, 3, 4, 5];

// This is the function
const compare = (a1, a2) =>
  a1.length == a2.length &&
  a1.every(
    (element, index) => element === a2[index]
  );

// Example
console.log(compare(array1, array2));

// Sorted
console.log(compare(array1.sort(), array2.sort()));

[Nicolas Hervé]

You forgot to compare array sizes.

compare([1], [1, 2]); // true

[virtualghostmck]

If a2 has more elements than a1, then this algo fails.

[Kavindu Santhusa]

Thanks, Now it's upadated.

[Shuvo]

Yes but we are talking about the simple approach here

[Shuvo]

And your function has some drawbacks also

[𒎏Wii 🏳️‍⚧️]

Whether that's a drawback or by design is questionable; whether two identical objects should be treated as actually the same completely depends on your problem domain.

[Shuvo]

agreed 100%

[Kavindu Santhusa]

In js your Objects are not equal
If your arrays are

const num5 = { num: 5 };
const array1 = [1, 2, 3, 4, num5];
const array2 = [1, 2, 3, 4, num5];

It works.

[Teerasak Vichadee]

I think it would be greate if we sort the array inside compare function, so we can sure that it will compare with the right position

[Marco Steinke]

This was the solution for a subtask of a problem I recently solved in a project.

I compared two vectors element-wise and counted the amount of distinct elements

getTotalDifference(anotherVector) {
    let diff = 0;
    this.values.forEach((e,i) => { return (this.values[i] != anotherVector[i]) ? diff++ : diff = diff; })
    return diff;
}