LeetCode-in-All

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

impl Solution {
    pub fn count_bits(num: i32) -> Vec<i32> {
        let mut result = vec![0; (num + 1) as usize];
        let mut border_pos = 1;
        let mut incr_pos = 1;
        
        for i in 1..=num {
            // When we reach a power of 2, reset border_pos and incr_pos
            if incr_pos == border_pos {
                result[i as usize] = 1;
                incr_pos = 1;
                border_pos = i;
            } else {
                result[i as usize] = 1 + result[incr_pos as usize];
                incr_pos += 1;
            }
        }
        
        result
    }
}

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