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Decimal to binary conversion without using arithmetic operators
• Difficulty Level : Medium
• Last Updated : 06 Apr, 2021

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```

## 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``        ``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.
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