338. Counting Bits

Problem Description

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++)?

Python Solution

class Solution:
    def countBits(self, n: int) -> List[int]:
        ans = [0] * (n + 1)
        for i in range(1, n + 1):
            # i >> 1 is i/2, so we can use previously calculated result
            # i & 1 is checking if i is odd (1) or even (0)
            ans[i] = ans[i >> 1] + (i & 1)
        return ans

Java Solution

class Solution {
    public int[] countBits(int n) {
        int[] ans = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            // i >> 1 is i/2, so we can use previously calculated result
            // i & 1 is checking if i is odd (1) or even (0)
            ans[i] = ans[i >> 1] + (i & 1);
        }
        return ans;
    }
}

C++ Solution

class Solution {
public:
    vector countBits(int n) {
        vector ans(n + 1);
        for (int i = 1; i <= n; i++) {
            // i >> 1 is i/2, so we can use previously calculated result
            // i & 1 is checking if i is odd (1) or even (0)
            ans[i] = ans[i >> 1] + (i & 1);
        }
        return ans;
    }
};

JavaScript Solution

/**
 * @param {number} n
 * @return {number[]}
 */
var countBits = function(n) {
    const ans = new Array(n + 1).fill(0);
    for (let i = 1; i <= n; i++) {
        // i >> 1 is i/2, so we can use previously calculated result
        // i & 1 is checking if i is odd (1) or even (0)
        ans[i] = ans[i >> 1] + (i & 1);
    }
    return ans;
};

C# Solution

public class Solution {
    public int[] CountBits(int n) {
        int[] ans = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            // i >> 1 is i/2, so we can use previously calculated result
            // i & 1 is checking if i is odd (1) or even (0)
            ans[i] = ans[i >> 1] + (i & 1);
        }
        return ans;
    }
}