Program to convert a Binary Number to Hexa-Decimal Number

Given a Binary Number, the task is to convert this Binary number to its equivalent Hexa-Decimal Number.
Examples:

Input: 100000101111 
Output: 82F 
Explanation:
Dividing the number into chunks of 4, it becomes 1000 0010 1111. 
Here, 1000 is equivalent to 8, 
0010 is equivalent to 2 and 
1111 is equivalent to F
Therefore, the number becomes 82F
Input:10101101 
Output:AD 
Explanation:
Dividing the number into chunks of 4, it becomes 1010 1101. 
Here, 1010 is equivalent to A and 
1101 is equivalent to D. 
Therefore, the number becomes AD. 

Approach:

1. Divide the given Binary number into chunks of 4, and start computing its equivalent HexaDecimal form.
2. Store this number formed in a vector.
3. Repeat the process for all the digits of the given Binary number.
4. Print the numbers stored in the vector in reverse order.

Below is the implementation of the above approach:

C++

// C++ code to convert Binary to its
// HexaDecimal number(base 16).

// Including Header Files
#include <bits/stdc++.h>
using namespace std;

// Function to convert
// Binary to HexaDecimal
void bToHexaDecimal(string s)
{
    int len = s.length(), check = 0;
    int num = 0, sum = 0, mul = 1;
    vector<char> ans;

    // Iterating through
    // the bits backwards
    for (int i = len - 1; i >= 0; i--) {
        sum += (s[i] - '0') * mul;
        mul *= 2;
        check++;

        // Computing the HexaDecimal
        // Number formed so far
        // and storing it in a vector.
        if (check == 4 || i == 0) {
            if (sum <= 9)
                ans.push_back(sum + '0');
            else
                ans.push_back(sum + 55);

            // Reinitializing all
            // variables for next group.
            check = 0;
            sum = 0;
            mul = 1;
        }
    }

    len = ans.size();

    // Printing the Hexadecimal
    // number formed so far.
    for (int i = len - 1; i >= 0; i--)
        cout << ans[i];
}

// Driver Code
int main()
{
    string s = "100000101111";

    // Function Call
    bToHexaDecimal(s);

    return 0;
}

Java

// Java code to convert BCD to its
// HexaDecimal number(base 16).
// Including Header Files
import java.util.*;
class GFG{

// Function to convert
// BCD to HexaDecimal
static void bcdToHexaDecimal(char []s)
{
  int len = s.length, check = 0;
  int num = 0, sum = 0, mul = 1;
  Vector<Character> ans =
         new Vector<Character>();

  // Iterating through
  // the bits backwards
  for (int i = len - 1; i >= 0; i--)
  {
    sum += (s[i] - '0') * mul;
    mul *= 2;
    check++;

    // Computing the HexaDecimal
    // Number formed so far
    // and storing it in a vector.
    if (check == 4 || i == 0)
    {
      if (sum <= 9)
        ans.add((char) (sum + '0'));
      else
        ans.add((char) (sum + 55));

      // Reinitializing all
      // variables for next group.
      check = 0;
      sum = 0;
      mul = 1;
    }
  }

  len = ans.size();

  // Printing the Hexadecimal
  // number formed so far.
  for (int i = len - 1; i >= 0; i--)
    System.out.print(ans.get(i));
}

// Driver Code
public static void main(String[] args)
{
  String s = "100000101111";

  // Function Call
  bcdToHexaDecimal(s.toCharArray());
}
}

// This code is contributed by Princi Singh

Python3

# Python3 code to convert BCD to its
# hexadecimal number(base 16).

# Function to convert BCD to hexadecimal
def bcdToHexaDecimal(s):

    len1 = len(s)
    check = 0
    num = 0
    sum = 0
    mul = 1
    ans = []

    # Iterating through the bits backwards
    i = len1 - 1

    while(i >= 0):
        sum += (ord(s[i]) - ord('0')) * mul
        mul *= 2
        check += 1

        # Computing the hexadecimal number formed
        # so far and storing it in a vector.
        if (check == 4 or i == 0):

            if (sum <= 9):
                ans.append(chr(sum + ord('0')))
            else:
                ans.append(chr(sum + 55));

            # Reinitializing all variables
            # for next group.
            check = 0
            sum = 0
            mul = 1

        i -= 1

    len1 = len(ans)

    # Printing the hexadecimal
    # number formed so far.
    i = len1 - 1

    while(i >= 0):
        print(ans[i], end = "")
        i -= 1

# Driver Code
if __name__ == '__main__':

    s = "100000101111"

    # Function Call
    bcdToHexaDecimal(s)

# This code is contributed by Samarth

C#

// C# code to convert BCD to its
// HexaDecimal number(base 16).
// Including Header Files
using System;
using System.Collections.Generic;
class GFG{

// Function to convert
// BCD to HexaDecimal
static void bcdToHexaDecimal(char []s)
{
  int len = s.Length, check = 0;
  int num = 0, sum = 0, mul = 1;
  List<char> ans =
       new List<char>();

  // Iterating through
  // the bits backwards
  for (int i = len - 1; i >= 0; i--)
  {
    sum += (s[i] - '0') * mul;
    mul *= 2;
    check++;

    // Computing the HexaDecimal
    // Number formed so far
    // and storing it in a vector.
    if (check == 4 || i == 0)
    {
      if (sum <= 9)
        ans.Add((char) (sum + '0'));
      else
        ans.Add((char) (sum + 55));

      // Reinitializing all
      // variables for next group.
      check = 0;
      sum = 0;
      mul = 1;
    }
  }

  len = ans.Count;

  // Printing the Hexadecimal
  // number formed so far.
  for (int i = len - 1; i >= 0; i--)
    Console.Write(ans[i]);
}

// Driver Code
public static void Main(String[] args)
{
  String s = "100000101111";

  // Function Call
  bcdToHexaDecimal(s.ToCharArray());
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// Javascript code to convert Binary to its
// HexaDecimal number(base 16).

// Including Header Files


// Function to convert
// Binary to HexaDecimal
function bToHexaDecimal(s) {
    let len = s.length, check = 0;
    let num = 0, sum = 0, mul = 1;
    let ans = new Array();

    // Iterating through
    // the bits backwards
    for (let i = len - 1; i >= 0; i--) {
        sum += (s[i].charCodeAt(0) - '0'.charCodeAt(0)) * mul;
        mul *= 2;
        check++;

        // Computing the HexaDecimal
        // Number formed so far
        // and storing it in a vector.
        if (check == 4 || i == 0) {
            if (sum <= 9)
                ans.push(String.fromCharCode(sum + '0'.charCodeAt(0)));
            else
                ans.push(String.fromCharCode(sum + 55));

            // Reinitializing all
            // variables for next group.
            check = 0;
            sum = 0;
            mul = 1;
        }
    }

    len = ans.length;

    // Printing the Hexadecimal
    // number formed so far.
    for (let i = len - 1; i >= 0; i--)
        document.write(ans[i]);
}

// Driver Code

let s = "100000101111";

// Function Call
bToHexaDecimal(s);


// This code is contributed by _saurabh_jaiswal
</script>

Output:

82F

Approach#2: Using dictionary

Create a dictionary mapping each 4-bit binary number to its corresponding hexadecimal equivalent. Pad the input binary number with zeros at the beginning to make its length a multiple of 4. Split the padded binary number into 4-bit chunks and map each chunk to its corresponding hexadecimal equivalent.the hexadecimal equivalents to get the final hexadecimal number.

Algorithm

1. Create a dictionary hex_dict with keys as 4-bit binary numbers and values as their corresponding hexadecimal equivalents.
2. Pad the input binary number with zeros at the beginning to make its length a multiple of 4.
3. Initialize an empty string hex_num.
4. Loop through the binary number in steps of 4 bits.
5. For each 4-bit chunk, map it to its corresponding hexadecimal equivalent using the hex_dict and append the result to hex_num.
6. Return the final hexadecimal number.

Python3

binary_num = "10101101"

hex_dict = {
    "0000": "0",
    "0001": "1",
    "0010": "2",
    "0011": "3",
    "0100": "4",
    "0101": "5",
    "0110": "6",
    "0111": "7",
    "1000": "8",
    "1001": "9",
    "1010": "A",
    "1011": "B",
    "1100": "C",
    "1101": "D",
    "1110": "E",
    "1111": "F"
}

hex_num = ""
n = len(binary_num)

# Pad the binary number with zeros at the beginning to make its length a multiple of 4
if n % 4 != 0:
    binary_num = "0" * (4 - n % 4) + binary_num
    n = len(binary_num)

# Convert the binary number to hexadecimal
for i in range(0, n, 4):
    hex_chunk = binary_num[i:i+4]
    hex_num += hex_dict[hex_chunk]

print(hex_num)

Output

AD

Time complexity: O(n), where n is the length of the input binary number. The for loop runs n/4 times, each time accessing a dictionary value in constant time.

Space complexity: O(1). The space used is constant, as only a dictionary and a string are used to store the input and output values.