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Decimal to binary conversion without using arithmetic operators

• Difficulty Level : Medium
• Last Updated : 16 Jan, 2023

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.

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