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338. Counting Bits

Easy

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

function countBits(n: number): number[] {
    const bits: number[] = [0]
    for (let i = 1; i <= n; i++) {
        bits[i] = bits[Math.floor(i / 2)] + (i % 2)
    }
    return bits
}

export { countBits }

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