Binary numbers of N digits

Namachila November 19, 2025

0 1 4 minutes read

Given a positive integer number N. The task is to generate all the binary numbers of N digits. These binary numbers should be in ascending order.

Examples:

Input: 2
Output:
00
01
10
11
Explanation: These 4 are the only binary numbers having 2 digits.

Input: 3
Output:
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1

Approach: For any digit length N, there will be 2^N binary numbers.

Therefore traverse from 0 to 2^N and convert every number to binary.
Store each number and print it at the end.

Below is the implementation of the above approach.

C++

// C++ code to implement above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to convert number
// to binary of N bits
vector<int> convertToBinary(int num, 
                            int length)
{
    // Vector to store the number
    vector<int> bits(length, 0);
    if (num == 0) {
        return bits;
    }
 
    int i = length - 1;
    while (num != 0) {
        bits[i--] = (num % 2);
 
        // Integer division
        // gives quotient
        num = num / 2;
    }
    return bits;
}
 
// Function to generate all
// N bit binary numbers
vector<vector<int> > getAllBinary(int n)
{
    vector<vector<int> > binary_nos;
 
    // Loop to generate the binary numbers
    for (int i = 0; i < pow(2, n); i++) {
        vector<int> bits = 
            convertToBinary(i, n);
        binary_nos.push_back(bits);
    }
 
    return binary_nos;
}
 
// Driver code
int main()
{
    int N = 3;
    vector<vector<int> > binary_nos = 
        getAllBinary(N);
    for (int i = 0; i < binary_nos.size(); 
           i++) {
        for (int j = 0; 
                j < binary_nos[i].size(); j++)
            cout << binary_nos[i][j];
        cout << endl;
    }
    return 0;
}

Java

// Java code for the above approach
import java.io.*;
 
class GFG {
 
  // Function to convert number
  // to binary of N bits
  static int[] convertToBinary(int num, 
                               int length)
  {
 
    // Vector to store the number
    int[] bits = new int[length];
    if (num == 0) {
      return bits;
    }
 
    int i = length - 1;
    while (num != 0) {
      bits[i--] = (num % 2);
 
      // Integer division
      // gives quotient
      num = num / 2;
    }
    return bits;
  }
 
  // Function to generate all
  // N bit binary numbers
  static int[][] getAllBinary(int n)
  {
    int[][] binary_nos = new int[(int)Math.pow(2,n)][];
    int k = 0;
     
    // Loop to generate the binary numbers
    for (int i = 0; i < Math.pow(2, n); i++) {
      int[] bits = convertToBinary(i, n);
      binary_nos[k++]= bits;
    }
 
    return binary_nos;
  }
 
  // Driver code
  public static void main (String[] args)
  {
    int N = 3;
    int[][] binary_nos = getAllBinary(N);
    for (int i = 0; i < binary_nos.length; i++) {
      for (int j = 0; j < binary_nos[i].length; j++)
        System.out.print(binary_nos[i][j]);
      System.out.println();
    }
 
  }
};
 
// This code is contributed by Potta Lokesh

Python3

# Python 3 code to implement above approach
 
# Function to convert number
# to binary of N bits
def convertToBinary(num,
                    length):
 
    # Vector to store the number
    bits = [0]*(length)
    if (num == 0):
        return bits
 
    i = length - 1
    while (num != 0):
        bits[i] = (num % 2)
        i -= 1
 
        # Integer division
        # gives quotient
        num = num // 2
 
    return bits
 
# Function to generate all
# N bit binary numbers
def getAllBinary(n):
 
    binary_nos = []
 
    # Loop to generate the binary numbers
    for i in range(pow(2, n)):
        bits = convertToBinary(i, n)
        binary_nos.append(bits)
 
    return binary_nos
 
# Driver code
if __name__ == "__main__":
 
    N = 3
    binary_nos = getAllBinary(N)
    for i in range(len(binary_nos)):
        for j in range(len(binary_nos[i])):
            print(binary_nos[i][j], end="")
        print()
 
        # This code is contributed by ukasp.

C#

// C# code for the above approach
using System;
using System.Collections.Generic;
public class GFG 
{
 
  // Function to convert number
  // to binary of N bits
  static List<int> convertToBinary(int num, int length) {
 
    // List to store the number
    List<int> bits = new List<int>();
    if (num == 0) {
      bits.Add(0);
      bits.Add(0);
      bits.Add(0);
      return bits;
    }
 
    int i = length - 1;
    while (num != 0) {
      bits.Add(num % 2);
 
      // int division
      // gives quotient
      num = num / 2;
    }
    while(bits.Count<3)
      bits.Add(0);
    return bits;
  }
 
  // Function to generate all
  // N bit binary numbers
  static List<List<int>> getAllBinary(int n) {
    List<List<int>> binary_nos = new List<List<int>>();
 
    // Loop to generate the binary numbers
    for (int i = 0; i < Math.Pow(2, n); i++) {
      List<int> bits = convertToBinary(i, n);
      binary_nos.Add(bits);
    }
 
    return binary_nos;
  }
 
  // Driver code
  public static void Main(String[] args) {
    int N = 3;
    List<List<int>> binary_nos = getAllBinary(N);
    foreach(var st in binary_nos){
      foreach(var s in st){
        Console.Write(s);
      }
      Console.WriteLine();
    }
  }
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
    // JavaScript code to implement above approach
 
    // Function to convert number
    // to binary of N bits
    const convertToBinary = (num, length) => {
        // Vector to store the number
        let bits = new Array(length).fill(0);
        if (num == 0) {
            return bits;
        }
 
        let i = length - 1;
        while (num != 0) {
            bits[i--] = (num % 2);
 
            // Integer division
            // gives quotient
            num = parseInt(num / 2);
        }
        return bits;
    }
 
    // Function to generate all
    // N bit binary numbers
    const getAllBinary = (n) => {
        let binary_nos = [];
 
        // Loop to generate the binary numbers
        for (let i = 0; i < parseInt(Math.pow(2, n)); i++) {
            let bits = convertToBinary(i, n);
            binary_nos.push(bits);
        }
 
        return binary_nos;
    }
 
    // Driver code
 
    let N = 3;
    let binary_nos = getAllBinary(N);
    for (let i = 0; i < binary_nos.length;
        i++) {
        for (let j = 0;
            j < binary_nos[i].length; j++)
            document.write(binary_nos[i][j]);
        document.write("<br/>");
    }
 
    // This code is contributed by rakeshsahni
 
</script>

Output

Time Complexity: O(2^N)
Auxiliary Space: O(2^N)

Recommended

Solve DSA problems on GfG Practice.

Solve Problems

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!