[Leetcode] 338. Counting Bits - JavaScript

This algorithm problem is from leetcode.

Problem

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 = 2 Output: [0,1,1] Explanation: 0 –> 0 1 –> 1 2 –> 10 Example 2:

Input: n = 5 Output: [0,1,1,2,1,2] Explanation: 0 –> 0 1 –> 1 2 –> 10 3 –> 11 4 –> 100 5 –> 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++)?

Submission 1 - Shifting Bits

Runtime 81 ms | Memory 49.4 MB
I used bitwise >> operator and &.

Instead of counting number of 1’s in given i one by one, I could use number of 1’s in previously calculated number that is less than i and has the common 1’s with given i. This picked number is i/2, which can be calculated by ` i » 1. Since i/2 or i » 1 will have the same number of 1's except the last bit, I add the last bit (1 or 0) of i to the number of 1's of i/2 by doing i & 1`.

var countBits = function(n) {
    const outputArr = [];
    outputArr[0] = 0; 

    for (let i =1; i <= n; i++) {
        outputArr[i] = outputArr[i >> 1]  + (i & 1);
    }
    return outputArr;
};

Submission 2

Runtime 66 ms | Memory 48.4 MB This submission was faster than the previous one.

Instead of counting number of 1’s in given i one by one, I could use the number of 1’s of the closest pre-calculated number that has the common number of 1’s with i. By running i & (i-1), I could “filter” out the 1’s of the closest number that has the maximum number of common 1’s with i. i & (i-1) could result in i-1 or some other previous number that doesn’t contain 1’s of i-1 not in i (for example, 1011 & 1100 will result in 1000. 1000 doesn’t contain 0011 that is in i-1=1011, but not in i=1100). After that, I just need to add i’s additional 1 bit to the number of 1’s of the closest number.

/**
 * @param {number} n
 * @return {number[]}
 */
var countBits = function(n) {
    const outputArr = [];
    outputArr[0] = 0; 

    for (let i =1; i <= n; i++) {
        outputArr[i] = outputArr[i & (i-1)] + 1;
    }
    return outputArr;
};