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# Decimal to binary conversion without using arithmetic operators

Find the binary equivalent of the given non-negative number n without using arithmetic operators.

Examples:

Input : n = 10
Output : 1010

Input : n = 38
Output : 100110

Note that + in below algorithm/program is used for concatenation purpose.
Algorithm:

```decToBin(n)
if n == 0
return "0"
Declare bin = ""
Declare ch
while n > 0
if (n & 1) == 0
ch = '0'
else
ch = '1'
bin = ch + bin
n = n >> 1
return bin```

Below is the implementation of above approach:

## C++

 `// C++ implementation of decimal to binary conversion``// without using arithmetic operators``#include ` `using` `namespace` `std;` `// function for decimal to binary conversion``// without using arithmetic operators``string decToBin(``int` `n)``{``    ``if` `(n == 0)``        ``return` `"0"``;``    ` `    ``// to store the binary equivalent of decimal``    ``string bin = ``""``;   ``    ``while` `(n > 0)   ``    ``{``        ``// to get the last binary digit of the number 'n'``        ``// and accumulate it at the beginning of 'bin'``        ``bin = ((n & 1) == 0 ? ``'0'` `: ``'1'``) + bin;``        ` `        ``// right shift 'n' by 1``        ``n >>= 1;``    ``}``    ` `    ``// required binary number``    ``return` `bin;``}` `// Driver program to test above``int` `main()``{``    ``int` `n = 38;``    ``cout << decToBin(n);``    ``return` `0;``}`

## Java

 `// Java implementation of decimal``// to binary conversion without``// using arithmetic operators``import` `java.io.*;` `class` `GFG {``    ` `    ``// function for decimal to``    ``// binary conversion without``    ``// using arithmetic operators``    ``static` `String decToBin(``int` `n)``    ``{``        ``if` `(n == ``0``)``            ``return` `"0"``;``    ` `        ``// to store the binary``        ``// equivalent of decimal``        ``String bin = ``""``;``        ` `        ``while` `(n > ``0``)``        ``{``            ``// to get the last binary digit``            ``// of the number 'n' and accumulate``            ``// it at the beginning of 'bin'``            ``bin = ((n & ``1``) == ``0` `? ``'0'` `: ``'1'``) + bin;``            ` `            ``// right shift 'n' by 1``            ``n >>= ``1``;``        ``}``        ` `        ``// required binary number``        ``return` `bin;``    ``}` `    ``// Driver program to test above``    ``public` `static` `void` `main (String[] args) {` `    ``int` `n = ``38``;``    ``System.out.println(decToBin(n));``    ``}``}` `// This code is contributed by vt_m`

## Python3

 `# Python3 implementation of``# decimal to binary conversion``# without using arithmetic operators` `# function for decimal to``# binary conversion without``# using arithmetic operators``def` `decToBin(n):``    ``if``(n ``=``=` `0``):``        ``return` `"0"``;``        ` `    ``# to store the binary``    ``# equivalent of decimal``    ``bin` `=` `"";``    ` `    ``while` `(n > ``0``):``        ` `        ``# to get the last binary``        ``# digit of the number 'n'``        ``# and accumulate it at``        ``# the beginning of 'bin'``        ``if` `(n & ``1` `=``=` `0``):``            ``bin` `=` `'0'` `+` `bin``;``        ``else``:``            ``bin` `=` `'1'` `+` `bin``;``        ` `        ``# right shift 'n' by 1``        ``# It also gives n//2 .``        ``n ``=` `n >> ``1``;``    ` `    ``# required binary number``    ``return` `bin``;` `# Driver Code``n ``=` `38``;``print``(decToBin(n));` `# This code is contributed``# by mits`

## C#

 `// C# implementation of decimal``// to binary conversion without``// using arithmetic operators``using` `System;` `class` `GFG {``    ` `    ``// function for decimal to``    ``// binary conversion without``    ``// using arithmetic operators``    ``static` `String decToBin(``int` `n)``    ``{``        ``if` `(n == 0)``            ``return` `"0"``;` `        ``// to store the binary``        ``// equivalent of decimal``        ``String bin = ``""``;` `        ``while` `(n > 0) {``            ` `            ``// to get the last binary digit``            ``// of the number 'n' and accumulate``            ``// it at the beginning of 'bin'``            ``bin = ((n & 1) == 0 ? ``'0'` `: ``'1'``) + bin;` `            ``// right shift 'n' by 1``            ``n >>= 1;``        ``}` `        ``// required binary number``        ``return` `bin;``    ``}` `    ``// Driver program to test above``    ``public` `static` `void` `Main()``    ``{` `        ``int` `n = 38;``        ``Console.WriteLine(decToBin(n));``    ``}``}` `// This code is contributed by Sam007`

## PHP

 ` 0)``    ``{``        ``// to get the last binary``        ``// digit of the number 'n'``        ``// and accumulate it at``        ``// the beginning of 'bin'``        ``\$bin` `= ((``\$n` `& 1) == 0 ?``                          ``'0'` `: ``'1'``) . ``\$bin``;``        ` `        ``// right shift 'n' by 1``        ``\$n` `>>= 1;``    ``}``    ` `    ``// required binary number``    ``return` `\$bin``;``}` `// Driver Code``\$n` `= 38;``echo` `decToBin(``\$n``);` `// This code is contributed``// by mits``?>`

## Javascript

 ``

Output:

`100110`

Time complexity: O(num), where num is the number of bits in the binary representation of n.
Auxiliary space: O(num), for using extra space for string bin.

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

METHOD 2:Using format()

APPROACH:

This code converts a decimal number to binary using the built-in format() function in Python. The function takes two arguments: the first is the number to be converted, and the second is the format specifier ‘b’, which tells the function to convert the number to binary.

ALGORITHM:

1. Take the decimal number as input.
2. Convert the decimal number to binary using the format() function with the format specifier ‘b’.
3. Store the result in a variable.
4. Print the variable.

## Python3

 `n ``=` `38``binary ``=` `format``(n, ``'b'``)``print``(f``"The binary representation of {n} is: {binary}"``)` `n ``=` `10``binary ``=` `format``(n, ``'b'``)``print``(f``"The binary representation of {n} is: {binary}"``)`

Output

```The binary representation of 38 is: 100110
The binary representation of 10 is: 1010
```

Time complexity: O(log n), where n is the decimal number, because the number of iterations required in the format() function depends on the number of bits required to represent the number in binary, which is log2(n).

Space complexity: O(log n), because the space required to store the binary representation of the number in the variable also depends on the number of bits required to represent the number in binary, which is log2(n).

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