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Binary representation of a given number

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Write a program to print Binary representation of a given number. 

Recommended Practice

Source: Microsoft Interview Set-3 

Method 1: Iterative 
For any number, we can check whether its ‘i’th bit is 0(OFF) or 1(ON) by bitwise ANDing it with “2^i” (2 raise to i). 

1) Let us take number 'NUM' and we want to check whether it's 0th bit is ON or OFF    
bit = 2 ^ 0 (0th bit)
if NUM & bit >= 1 means 0th bit is ON else 0th bit is OFF
2) Similarly if we want to check whether 5th bit is ON or OFF
bit = 2 ^ 5 (5th bit)
if NUM & bit >= 1 means its 5th bit is ON else 5th bit is OFF.

Let us take unsigned integer (32 bit), which consist of 0-31 bits. To print binary representation of unsigned integer, start from 31th bit, check whether 31th bit is ON or OFF, if it is ON print “1” else print “0”. Now check whether 30th bit is ON or OFF, if it is ON print “1” else print “0”, do this for all bits from 31 to 0, finally we will get binary representation of number.

C++
// C++ Program for the binary
// representation of a given number
#include <bits/stdc++.h>
using namespace std;

void bin(long n)
{
    long i;
    cout << "0";
    for (i = 1 << 31; i > 0; i = i / 2) {
        if ((n & i) != 0) {
            cout << "1";
        }
        else {
            cout << "0";
        }
    }
}

// Driver Code
int main(void)
{
    bin(7);
    cout << endl;
    bin(4);
}
C
#include<stdio.h>
void bin(unsigned n)
{
    unsigned i;
    for (i = 1 << 31; i > 0; i = i / 2)
        (n & i) ? printf("1") : printf("0");
}

int main(void)
{
    bin(7);
    printf("\n");
    bin(4);
}
Java
public class GFG
{
  static void bin(long n)
  {
    long i;
    System.out.print("0");
    for (i = 1 << 31; i > 0; i = i / 2)
    {
      if((n & i) != 0)
      {
        System.out.print("1");
      }
      else
      {
        System.out.print("0");
      }
    }
  }

  // Driver code
  public static void main(String[] args)
  {
    bin(7);
    System.out.println();
    bin(4);
  }
}

// This code is contributed by divyesh072019.
C#
using System;

public class GFG{
    static void bin(long n)
    {
        long i;
        Console.Write("0");
        for (i = 1 << 31; i > 0; i = i / 2)
        {
            if((n & i) != 0)
            {
                Console.Write("1");
            }
            else
            {
                Console.Write("0");
            }
        }
    }
    // Driver code
    static public void Main (){
        bin(7);
        Console.WriteLine();
        bin(4);
    }
}
// This code is contributed by avanitrachhadiya2155
Javascript
<script>
// JavaScript Program for the binary
// representation of a given number
    function bin(n)
    {
        let i;
        document.write("0");
        for (i = 1 << 30; i > 0; i = Math.floor(i/2))
        {
            if((n & i) != 0)
            {
                document.write("1");
            }
            else
            {
                document.write("0");
            }
        }
    }
    
    bin(7);
    document.write("<br>");
    bin(4);
    
    
    // This code is contributed by rag2127
</script>
Python3
def bin(n):

    i = 1 << 31
    while(i > 0):

        if((n & i) != 0):

            print("1", end="")

        else:
            print("0", end="")

        i = i // 2


bin(7)
print()
bin(4)

# This code is contributed by divyeshrabadiya07.

Output
0
0

Time Complexity: O(1)
Auxiliary Space: O(1)

Method 2: Recursive 
Following is recursive method to print binary representation of ‘NUM’. 

step 1) if NUM > 1
a) push NUM on stack
b) recursively call function with 'NUM / 2'
step 2)
a) pop NUM from stack, divide it by 2 and print it's remainder.
C++
// C++ Program for the binary
// representation of a given number
#include <bits/stdc++.h>
using namespace std;

void bin(unsigned n)
{
    /* step 1 */
    if (n > 1)
        bin(n / 2);

    /* step 2 */
    cout << n % 2;
}

// Driver Code
int main(void)
{
    bin(7);
    cout << endl;
    bin(4);
}
C
// C Program for the binary
// representation of a given number
void bin(unsigned n)
{
    /* step 1 */
    if (n > 1)
        bin(n / 2);

    /* step 2 */
    printf("%d", n % 2);
}

// Driver Code
int main(void)
{
    bin(7);
    printf("\n");
    bin(4);
}
Java
// Java Program for the binary
// representation of a given number
class GFG {
    static void bin(int n)
    {
        /* step 1 */
        if (n > 1)
            bin(n / 2);

        /* step 2 */
        System.out.print(n % 2);
    }

    // Driver code
    public static void main(String[] args)
    {
        bin(7);
        System.out.println();
        bin(4);
    }
}

// This code is contributed
// by ChitraNayal
C#
// C# Program for the binary
// representation of a given number
using System;

class GFG {

    static void bin(int n)
    {

        // step 1
        if (n > 1)
            bin(n / 2);

        // step 2
        Console.Write(n % 2);
    }

    // Driver code
    static public void Main()
    {
        bin(7);
        Console.WriteLine();
        bin(4);
    }
}

// This code is contributed ajit
Javascript
<script>

// Javascript program for the binary
// representation of a given number
function bin(n)
{
    
    // Step 1
    if (n > 1)
        bin(parseInt(n / 2, 10));

    // Step 2
    document.write(n % 2);
}

// Driver code
bin(7);
document.write("</br>");
bin(4);

// This code is contributed by divyeshrabadiya07

</script>
PHP
<?php
// PHP Program for the binary 
// representation of a given number
function bin($n)
{
    /* step 1 */
    if ($n > 1)
        bin($n/2);

    /* step 2 */
    echo ($n % 2);
}

// Driver code
bin(7);
echo ("\n");
bin(4);

// This code is contributed 
// by Shivi_Aggarwal
?>
Python3
# Python3 Program for the binary
# representation of a given number


def bin(n):

    if n > 1:
        bin(n//2)

    print(n % 2, end="")


# Driver Code
if __name__ == "__main__":

    bin(7)
    print()
    bin(4)

# This code is contributed by ANKITRAI1

Output
111
100

Time Complexity: O(log N)
Auxiliary Space: O(log N)

Method 3: Recursive using bitwise operator 
Steps to convert decimal number to its binary representation are given below: 

step 1: Check n > 0
step 2: Right shift the number by 1 bit and recursive function call
step 3: Print the bits of number
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to convert decimal
// to binary number
void bin(unsigned n)
{
    if (n > 1)
        bin(n >> 1);

    printf("%d", n & 1);
}

// Driver code
int main(void)
{
    bin(131);
    printf("\n");
    bin(3);
    return 0;
}
Java
// Java implementation of the approach

class GFG {

    // Function to convert decimal
    // to binary number
    static void bin(Integer n)
    {
        if (n > 1)
            bin(n >> 1);

        System.out.printf("%d", n & 1);
    }

    // Driver code
    public static void main(String[] args)
    {
        bin(131);
        System.out.printf("\n");
        bin(3);
    }
}
/*This code is contributed by PrinciRaj1992*/
C#
// C# implementation of above approach
using System;

public class GFG {

    // Function to convert decimal
    // to binary number
    static void bin(int n)
    {
        if (n > 1)
            bin(n >> 1);

        Console.Write(n & 1);
    }

    // Driver code
    public static void Main()
    {
        bin(131);
        Console.WriteLine();
        bin(3);
    }
}
/*This code is contributed by PrinciRaj1992*/
Javascript
<script>
// JavaScript implementation of the approach

// Function to convert decimal
// to binary number
function bin(n)
{
    if (n > 1)
        bin(n >> 1);

    document.write(n & 1);
}

// Driver code

    bin(131);
    document.write("<br>");
    bin(3);


// This code is contributed by Surbhi Tyagi.

</script>
PHP
<?php
// PHP implementation of the approach

// Function to convert decimal
// to binary number
function bin($n)
{
    if ($n > 1)
    bin($n>>1);
    
    echo ($n & 1);
}

// Driver code
bin(131);
echo "\n";
bin(3);

// This code is contributed
// by Akanksha Rai
Python3
# Python 3 implementation of above approach

# Function to convert decimal to
# binary number


def bin(n):

    if (n > 1):
        bin(n >> 1)
    print(n & 1, end="")


# Driver code
bin(131)
print()
bin(3)

# This code is contributed by PrinciRaj1992

Output
10000011
11

Time Complexity: O(log N)
Auxiliary Space: O(log N)

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Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
This article is compiled by Narendra Kangralkar



Last Updated : 12 Mar, 2024
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