338. Counting Bits

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++)?

Solution

/**
 * @param {number} n
 * @return {number[]}
 */
var countBits = function(n) {
    var res = Array(n + 1);
    var num = 2;
    var nextNum = 4;
    for (var i = 0; i <= n; i++) {
        if (i === 0) {
            res[i] = 0;
        } else if (i === 1) {
            res[i] = 1;
        } else {
            if (i === nextNum) {
                num = nextNum;
                nextNum *= 2;
            }
            res[i] = 1 + res[i - num];
        }
    }
    return res;
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).