Build Bit Count Array

Given an integer n, return an array containing the number of set bits (1s) in the binary representation of every integer in the range [0, n].

Examples :

Input: n = 3
Output: [0, 1, 1, 2]
Explanation: The numbers from 0 to 3 have 0, 1, 1, and 2 set bits respectively.

Input: n = 7
Output: [0, 1, 1, 2, 1, 2, 2, 3]
Explanation: The numbers from 0 to 7 have 0, 1, 1, 2, 1, 2, 2, and 3 set bits respectively.

Try It Yourself

Table of Content

[Naive Approach] Count Set Bits for Each Number - O(n log n) Time and O(1) Space
[Expected Approach] Using Dynamic Programming - O(n) Time and O(n) Space

[Naive Approach] Count Set Bits for Each Number - O(n log n) Time and O(1) Space

The idea is to iterate through all numbers from 0 to n and count the set bits in each number individually by checking every bit position. Store the count obtained for each number in the result array.

C++

#include <iostream>
#include <vector>
using namespace std;

// Function to count set bits for all numbers from 0 to n
vector<int> countBits(int n)
{

    // Stores count of set bits for each number
    vector<int> ans;

    // Process every number from 0 to n
    for (int i = 0; i <= n; i++)
    {
        int cnt = 0;
        int num = i;

        // Count set bits in current number
        while (num > 0)
        {
            cnt += (num & 1);
            num >>= 1;
        }

        ans.push_back(cnt);
    }

    return ans;
}

// Driver Code
int main()
{
    int n = 7;

    vector<int> res = countBits(n);

    cout << "[";

    for (int i = 0; i < res.size(); i++)
    {
        cout << res[i];

        if (i != res.size() - 1)
            cout << ", ";
    }

    cout << "]";

    return 0;
}

Java

import java.util.ArrayList;
import java.util.List;

// Function to count set bits for all numbers from 0 to n
public class GfG {
    public static List<Integer> countBits(int n)
    {

        // Stores count of set bits for each number
        List<Integer> ans = new ArrayList<>();

        // Process every number from 0 to n
        for (int i = 0; i <= n; i++) {
            int cnt = 0;
            int num = i;

            // Count set bits in current number
            while (num > 0) {
                cnt += (num & 1);
                num >>= 1;
            }

            ans.add(cnt);
        }

        return ans;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int n = 7;

        List<Integer> res = countBits(n);

        System.out.print("[");

        for (int i = 0; i < res.size(); i++) {
            System.out.print(res.get(i));

            if (i != res.size() - 1)
                System.out.print(", ");
        }

        System.out.print("]");
    }
}

Python

def countBits(n):
    
    # Stores count of set bits for each number
    ans = []

    # Process every number from 0 to n
    for i in range(n + 1):
        cnt = 0
        num = i

        # Count set bits in current number
        while num > 0:
            cnt += (num & 1)
            num >>= 1

        ans.append(cnt)

    return ans

# Driver Code
if __name__ == "__main__":
    n = 7

    res = countBits(n)

    print("[", end="")
    for i in range(len(res)):
        print(res[i], end="")
        if i != len(res) - 1:
            print(", ", end="")
    print("]")

using System;
using System.Collections.Generic;

// Function to count set bits for all numbers from 0 to n
public class GfG {
    public static List<int> countBits(int n)
    {

        // Stores count of set bits for each number
        List<int> ans = new List<int>();

        // Process every number from 0 to n
        for (int i = 0; i <= n; i++) {
            int cnt = 0;
            int num = i;

            // Count set bits in current number
            while (num > 0) {
                cnt += (num & 1);
                num >>= 1;
            }

            ans.Add(cnt);
        }

        return ans;
    }

    // Driver Code
    public static void Main()
    {
        int n = 7;

        List<int> res = countBits(n);

        Console.Write("[");

        for (int i = 0; i < res.Count; i++) {
            Console.Write(res[i]);

            if (i != res.Count - 1)
                Console.Write(", ");
        }

        Console.Write("]");
    }
}

JavaScript

// Function to count set bits for all numbers from 0 to n
function countBits(n) {

    // Stores count of set bits for each number
    let ans = [];

    // Process every number from 0 to n
    for (let i = 0; i <= n; i++) {
        let cnt = 0;
        let num = i;

        // Count set bits in current number
        while (num > 0) {
            cnt += (num & 1);
            num >>= 1;
        }

        ans.push(cnt);
    }

    return ans;
}

// Driver Code
let n = 7;
let res = countBits(n);
console.log(`[${res.join(', ')}]`);

Output

[0, 1, 1, 2, 1, 2, 2, 3]

[Expected Approach] Using Dynamic Programming - O(n) Time and O(n) Space

The idea is to use the count of set bits of a smaller number to compute the count for a larger number. For any number i, the count of set bits equals the count of set bits in i >> 1 plus 1 if the least significant bit of i is set. Using this relation, we can build the answer for all numbers from 0 to n in a bottom-up manner.

Let us understand with example:
Input: n = 7

Initialize bits = [0, 0, 0, 0, 0, 0, 0, 0], where bits[i] stores the count of set bits in i.
For i = 1, bits[1] = bits[0] + (1 & 1) = 0 + 1 = 1. Now, bits = [0, 1, 0, 0, 0, 0, 0, 0].
For i = 2 and i = 3, bits[2] = bits[1] + 0 = 1 and bits[3] = bits[1] + 1 = 2. Now, bits = [0, 1, 1, 2, 0, 0, 0, 0].
For i = 4, i = 5, and i = 6, we get bits[4] = 1, bits[5] = 2, and bits[6] = 2. Now, bits = [0, 1, 1, 2, 1, 2, 2, 0].
For i = 7, bits[7] = bits[3] + 1 = 2 + 1 = 3. The final array becomes [0, 1, 1, 2, 1, 2, 2, 3].

C++

#include <iostream>
#include <vector>
using namespace std;

vector<int> countBits(int n)
{

    // Stores the count of set bits for each number from 0 to n
    vector<int> bits(n + 1, 0);

    // Compute set bit count using dynamic programming
    for (int i = 1; i <= n; i++)
    {

        // Count of set bits in i equals the count of set bits
        // in i >> 1 plus 1 if the least significant bit is set
        bits[i] = bits[i >> 1] + (i & 1);
    }

    return bits;
}

// Driver Code
int main()
{
    int n = 7;

    vector<int> res = countBits(n);

    cout << "[";

    for (int i = 0; i < res.size(); i++)
    {
        cout << res[i];

        if (i != res.size() - 1)
            cout << ", ";
    }

    cout << "]";

    return 0;
}