The answer would be Option 1 because the length of the first array would be 3. first is a sparse array i.e. array with empty slots in between. The indices in sparse array itself doesn't exist rather than having undefined as value.

console.log(first.length); // 3

first[0] -- 1
first[1] -- empty slot
first[2] -- 3

Since the array is just an object then its representation would be
// first
{
    "0": 1,
    "2": 3
}

Now, when we use the spread operator to create a copy then it will iterate over the array as in i: 0 to n - 1 and set the value second[i] = first[i]. Since, when we access empty slots in an array then the returned value would be undefined, therefore, second[1] would be undefined. By spreading, we just created a dense array.

// second
{
    "0": 1,
    "1": undefined,
    "2": 3
}

Array.prototype.forEach iterates over the array and check if the indices exist in the array. If yes, then only callback will be invoked with index & value as arguments.

To know more --

https://github.com/v8/v8/blob/ca6e40d7ba853319c15196fef3f4536c8b3929fe/src/js/array.js#L1059

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/forEach

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What would be the output? (Based on Array prototype forEach)

@Yomesh Gupta

Option 1

Option 2

Option 3

Option 4