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# Convert a given Decimal number to its BCD representation

Given a decimal number N, the task is to convert N to it’s Binary Coded Decimal(BCD) form.
Examples:

Input: N = 12
Output: 0001 0000
Explanation:
Considering 4-bit concept:
1 in binary is 0001 and 2 in binary is 0010
So it’s equivalent BCD is 0001 0010.

Input: N = 10
Output: 0001 0000
Explanation:
Considering 4-bit concept:
1 in binary is 0001 and 0 in binary is 0000
So it’s equivalent BCD is 0001 0000.

Approach:

1. Reverse the digits of the given number N using the approach discussed in this article and stored the number in Rev.
2. Extract the digits of Rev and print the Binary form of the digit using bitset.
3. Repeat the above steps for each digit in Rev.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to convert Decimal to BCD` `void` `BCDConversion(``int` `n)` `{` `    ``// Base Case` `    ``if` `(n == 0) {` `        ``cout << ``"0000"``;` `        ``return``;` `    ``}`   `    ``// To store the reverse of n` `    ``int` `rev = 0;`   `    ``// Reversing the digits` `    ``while` `(n > 0) {` `        ``rev = rev * 10 + (n % 10);` `        ``n /= 10;` `    ``}`   `    ``// Iterate through all digits in rev` `    ``while` `(rev > 0) {`   `        ``// Find Binary for each digit` `        ``// using bitset` `        ``bitset<4> b(rev % 10);`   `        ``// Print the Binary conversion` `        ``// for current digit` `        ``cout << b << ``' '``;`   `        ``// Divide rev by 10 for next digit` `        ``rev /= 10;` `    ``}` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given Number` `    ``int` `N = 12;`   `    ``// Function Call` `    ``BCDConversion(N);` `    ``return` `0;` `}`

## Java

 `// Java program for the above approach ` `import` `java.util.*;`   `class` `Gfg` `{` `  `  `    ``// Function to convert Decimal to BCD` `    ``public` `static` `void` `BCDConversion(``int` `n)` `    ``{ ` `      `  `        ``// Base Case` `        ``if``(n == ``0``)` `        ``{` `            ``System.out.print(``"0000"``);` `        ``}` `      `  `        ``// To store the reverse of n` `        ``int` `rev = ``0``;` `      `  `        ``// Reversing the digits` `        ``while` `(n > ``0``) ` `        ``{` `            ``rev = rev * ``10` `+ (n % ``10``);` `            ``n /= ``10``;` `        ``}` `        `  `        ``// Iterate through all digits in rev` `        ``while``(rev > ``0``)` `        ``{` `          `  `            ``// Find Binary for each digit` `            ``// using bitset` `            ``String b = Integer.toBinaryString(rev % ``10``);` `            `  `            ``b = String.format(``"%04d"``, Integer.parseInt(b));` `            `  `              ``// Print the Binary conversion` `            ``// for current digit` `            ``System.out.print(b + ``" "``);` `          `  `            ``// Divide rev by 10 for next digit` `            ``rev /= ``10``;` `        ``}` `    ``}` `  `  `  ``// Driver code` `  ``public` `static` `void` `main(String []args)` `  ``{` `    `  `    ``// Given Number` `    ``int` `N = ``12``;` `    `  `    ``// Function Call` `    ``BCDConversion(N);` `  ``}` `}`   `// This code is contributed by avanitrachhadiya2155`

## Python3

 `# Python3 program for the above approach `   `# Function to convert Decimal to BCD ` `def` `BCDConversion(n) : `   `    ``# Base Case ` `    ``if` `(n ``=``=` `0``) : ` `        ``print``(``"0000"``)` `        ``return`   `    ``# To store the reverse of n ` `    ``rev ``=` `0`   `    ``# Reversing the digits ` `    ``while` `(n > ``0``) : ` `        ``rev ``=` `rev ``*` `10` `+` `(n ``%` `10``)` `        ``n ``=` `n ``/``/` `10`   `    ``# Iterate through all digits in rev ` `    ``while` `(rev > ``0``) : `   `        ``# Find Binary for each digit ` `        ``# using bitset ` `        ``b ``=` `str``(rev ``%` `10``)` `        `  `        ``# Print the Binary conversion ` `        ``# for current digit ` `        ``print``(``"{0:04b}"``.``format``(``int``(b, ``16``)), end ``=` `" "``) `   `        ``# Divide rev by 10 for next digit ` `        ``rev ``=` `rev ``/``/` `10`   `# Given Number ` `N ``=` `12`   `# Function Call ` `BCDConversion(N)`   `# This code is contributed by divyeshrabadiya07`

## C#

 `// C# program for the above approach ` `using` `System;` `using` `System.Collections.Generic; ` `class` `GFG {` `    `  `    ``// Function to convert Decimal to BCD` `    ``static` `void` `BCDConversion(``int` `n)` `    ``{` `        ``// Base Case` `        ``if` `(n == 0) {` `            ``Console.Write(``"0000"``);` `            ``return``;` `        ``}` `     `  `        ``// To store the reverse of n` `        ``int` `rev = 0;` `     `  `        ``// Reversing the digits` `        ``while` `(n > 0) {` `            ``rev = rev * 10 + (n % 10);` `            ``n /= 10;` `        ``}` `     `  `        ``// Iterate through all digits in rev` `        ``while` `(rev > 0) {` `     `  `            ``// Find Binary for each digit` `            ``// using bitset` `            ``string` `b = Convert.ToString(rev % 10, 2).PadLeft(4, ``'0'``);` `     `  `            ``// Print the Binary conversion` `            ``// for current digit` `            ``Console.Write(b + ``" "``);` `     `  `            ``// Divide rev by 10 for next digit` `            ``rev /= 10;` `        ``}` `    ``}`   `  ``static` `void` `Main() {` `      `  `    ``// Given Number` `    ``int` `N = 12;` ` `  `    ``// Function Call` `    ``BCDConversion(N);` `  ``}` `}`   `// This code is contributed divyesh072019`

## Javascript

 ``

Output:

`0001 0010`

Time Complexity: O(log10 N), where N is the given number.
Auxiliary Space: O(1) because constant space has been used

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