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 2N binary numbers.
- Therefore traverse from 0 to 2N 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 bitsvector<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 numbersvector<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 codeint 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 approachimport 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 bitsdef 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 numbersdef 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 codeif __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 approachusing 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
000 001 010 011 100 101 110 111
Time Complexity: O(2N)
Auxiliary Space: O(2N)
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