338. 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 = 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: #

impl Solution {
    pub fn count_bits(n: i32) -> Vec<i32> {
        let mut res = vec![0; n as usize + 1];

        for i in 1..=n {
            let i = i as usize;
            // let i_and_i_minus_1 = i & (i - 1);
            // res[i] = res[i_and_i_minus_1] + 1;
            res[i] = res[i >> 1] + (i & 1) as i32;
        }

        res
    }
}

... #