Counting Bits

Problem statement

Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1's in the binary representation of i.

Example 1:

Input: n = 2Output: [0,1,1]Explanation:0 --> 01 --> 12 --> 10

Example 2:

Input: n = 5Output: [0,1,1,2,1,2]Explanation:0 --> 01 --> 12 --> 103 --> 114 --> 1005 --> 101

Constraints:

• 0 <= n <= 105

Follow up:

• It is very easy to come up with a solution with a runtime of O(n log n). Can you do it in linear time O(n) and possibly in a single pass?
• Can you do it without using any built-in function (i.e., like __builtin_popcount in C++)?

My solution

/**
 * @param {number} n
 * @return {number[]}
 */
var countBits = function(n) {
    // function convertToBinary(number) {
    //     let num = number;
    //     let binary = (num % 2).toString();
    //     console.log("binary", binary)
    //     for (; num > 1; ) {
    //         num = parseInt(num / 2);
    //         console.log("num", num)
    //         binary =  (num % 2) + (binary);
    //         console.log("binary", binary)
    //     }
    // }
    
    const arr = Array.from({
        length: n + 1
    }, (_, index) => {
        const ones = index.toString(2).split("").filter(v => v === "1")
        // console.log(ones)
        return ones.length
    })
    
    return arr
};