README.md

338. Counting Bits

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.

Example 1:

Input: 2
Output: [0,1,1]

Example 2:

Input: 5
Output: [0,1,1,2,1,2]

Follow up:

• It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
• Space complexity should be O(n).
• Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

Solutions (Rust)

1. i % n (n = 2^{⌊log₂(n)⌋})

impl Solution {
    pub fn count_bits(num: i32) -> Vec<i32> {
        let mut n = 1;
        let mut result = vec![0];
        for i in 1..=num as usize{
            n *= (i / n);
            result.push(&result[i % n] + 1);
        }
        result
    }
}

2. i & (i - 1)

impl Solution {
    pub fn count_bits(num: i32) -> Vec<i32> {
        let mut result = vec![0];
        for i in 1..=num as usize{
            result.push(&result[i & (i - 1)] + 1);
        }
        result
    }
}