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Program to Convert BCD number into Decimal number

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Given a BCD (Binary Coded Decimal) number, the task is to convert the BCD number into its equivalent Decimal number

Examples:  

Input: BCD = 100000101000 
Output: 828 
Explanation: 
Dividing the number into chunks of 4, it becomes 1000 0010 1000
Here, 1000 is equivalent to 8 and 
0010 is equivalent to 2
So, the number becomes 828.

Input: BCD = 1001000 
Output: 48 
Explanation: 
Dividing the number into chunks of 4, it becomes 0100 1000
Here, 0100 is equivalent to 4 and 
1000 is equivalent to 8
So, the number becomes 48.  

Approach:  

  1. Iterate over all bits in given BCD numbers.
  2. Divide the given BCD number into chunks of 4, and start computing its equivalent Decimal number
  3. Store this number formed in a variable named sum
  4. Start framing a number from the digits stored in the sum in a variable num
  5. Reverse the number formed so far and return that number. 
     

Below is the implementation of the above approach.  

C++




// C++ code to convert BCD to its
// decimal number(base 10).
 
// Including Header Files
#include <bits/stdc++.h>
using namespace std;
 
// Function to convert BCD to Decimal
int bcdToDecimal(string s)
{
    int len = s.length(),
        check = 0, check0 = 0;
    int num = 0, sum = 0,
        mul = 1, rev = 0;
 
    // Iterating through the bits backwards
    for (int i = len - 1; i >= 0; i--) {
 
        // Forming the equivalent
        // digit(0 to 9)
        // from the group of 4.
        sum += (s[i] - '0') * mul;
        mul *= 2;
        check++;
 
        // Reinitialize all variables
        // and compute the number.
        if (check == 4 || i == 0) {
            if (sum == 0 && check0 == 0) {
                num = 1;
                check0 = 1;
            }
            else {
                // update the answer
                num = num * 10 + sum;
            }
 
            check = 0;
            sum = 0;
            mul = 1;
        }
    }
 
    // Reverse the number formed.
    while (num > 0) {
        rev = rev * 10 + (num % 10);
        num /= 10;
    }
 
    if (check0 == 1)
        return rev - 1;
 
    return rev;
}
 
// Driver Code
int main()
{
    string s = "100000101000";
 
    // Function Call
    cout << bcdToDecimal(s);
 
    return 0;
}


Java




// Java code to convert BCD to its
// decimal number(base 10).
// Including Header Files
import java.io.*;
import java.util.*;
 
class GFG {
     
// Function to convert BCD to Decimal
public static int bcdToDecimal(String s)
{
    int len = s.length();
    int check = 0, check0 = 0;
    int num = 0, sum = 0;
    int mul = 1, rev = 0;
 
    // Iterating through the bits backwards
    for(int i = len - 1; i >= 0; i--)
    {
 
       // Forming the equivalent
       // digit(0 to 9)
       // from the group of 4.
       sum += (s.charAt(i) - '0') * mul;
       mul *= 2;
       check++;
        
       // Reinitialize all variables
       // and compute the number.
       if (check == 4 || i == 0)
       {
           if (sum == 0 && check0 == 0)
           {
               num = 1;
               check0 = 1;
           }
           else
           {
 
               // Update the answer
               num = num * 10 + sum;
           }
            
           check = 0;
           sum = 0;
           mul = 1;
       }
    }
     
    // Reverse the number formed.
    while (num > 0)
    {
        rev = rev * 10 + (num % 10);
        num /= 10;
    }
 
    if (check0 == 1)
        return rev - 1;
 
    return rev;
}
 
// Driver code
public static void main(String[] args)
{
    String s = "100000101000";
 
    // Function Call
    System.out.println(bcdToDecimal(s));
}
}
 
// This code is contributed by coder001


Python3




# Python3 code to convert BCD to its
# decimal number(base 10).
 
# Function to convert BCD to Decimal
def bcdToDecimal(s):
 
    length = len(s);
    check = 0;
    check0 = 0;
    num = 0;
    sum = 0;
    mul = 1;
    rev = 0;
     
    # Iterating through the bits backwards
    for i in range(length - 1, -1, -1):
         
        # Forming the equivalent
        # digit(0 to 9)
        # from the group of 4.
        sum += (ord(s[i]) - ord('0')) * mul;
        mul *= 2;
        check += 1;
         
        # Reinitialize all variables
        # and compute the number
        if (check == 4 or i == 0):
            if (sum == 0 and check0 == 0):
                num = 1;
                check0 = 1;
                 
            else:
                 
                # Update the answer
                num = num * 10 + sum;
                 
            check = 0;
            sum = 0;
            mul = 1;
             
    # Reverse the number formed.
    while (num > 0):
        rev = rev * 10 + (num % 10);
        num //= 10;
         
    if (check0 == 1):
        return rev - 1;
         
    return rev;
 
# Driver Code
if __name__ == "__main__":
 
    s = "100000101000";
     
    # Function Call
    print(bcdToDecimal(s));
     
# This code is contributed by AnkitRai01


C#




// C# code to convert BCD to its
// decimal number(base 10).
// Including Header Files
using System;
 
class GFG
{
     
// Function to convert BCD to Decimal
public static int bcdToDecimal(String s)
{
    int len = s.Length;
    int check = 0, check0 = 0;
    int num = 0, sum = 0;
    int mul = 1, rev = 0;
 
    // Iterating through the bits backwards
    for(int i = len - 1; i >= 0; i--)
    {
 
        // Forming the equivalent
        // digit(0 to 9)
        // from the group of 4.
        sum += (s[i] - '0') * mul;
        mul *= 2;
        check++;
             
        // Reinitialize all variables
        // and compute the number.
        if (check == 4 || i == 0)
        {
            if (sum == 0 && check0 == 0)
            {
                num = 1;
                check0 = 1;
            }
            else
            {
     
                // Update the answer
                num = num * 10 + sum;
            }
             
            check = 0;
            sum = 0;
            mul = 1;
        }
    }
     
    // Reverse the number formed.
    while (num > 0)
    {
        rev = rev * 10 + (num % 10);
        num /= 10;
    }
 
    if (check0 == 1)
        return rev - 1;
 
    return rev;
}
 
// Driver code
public static void Main(String[] args)
{
    String s = "100000101000";
 
    // Function Call
    Console.WriteLine(bcdToDecimal(s));
}
}
 
// This code is contributed by 29AjayKumar


Javascript




<script>
 
// Javascript code to convert BCD to its
// decimal number(base 10).
// Including Header Files
 
// Function to convert BCD to Decimal
function bcdToDecimal(s)
{
    let len = s.length;
    let check = 0, check0 = 0;
    let num = 0, sum = 0;
    let mul = 1, rev = 0;
  
    // Iterating through the bits backwards
    for(let i = len - 1; i >= 0; i--)
    {
  
       // Forming the equivalent
       // digit(0 to 9)
       // from the group of 4.
       sum += (s[i] - '0') * mul;
       mul *= 2;
       check++;
         
       // Reinitialize all variables
       // and compute the number.
       if (check == 4 || i == 0)
       {
           if (sum == 0 && check0 == 0)
           {
               num = 1;
               check0 = 1;
           }
           else
           {
  
               // Update the answer
               num = num * 10 + sum;
           }
             
           check = 0;
           sum = 0;
           mul = 1;
       }
    }
      
    // Reverse the number formed.
    while (num > 0)
    {
        rev = rev * 10 + (num % 10);
        num = Math.floor(num / 10);
    }
  
    if (check0 == 1)
        return rev - 1;
  
    return rev;
}
 
// Driver Code
     
    let s = "100000101000";
  
    // Function Call
    document.write(bcdToDecimal(s.split('')));
       
</script>


Output

828








Time complexity: O(N) 
Auxiliary Space complexity: O(1) 

Method: Using bit manipulation method

The bcdToDecimal function takes a string s representing the BCD number as input and returns its decimal equivalent. It first initializes the num variable to 0. Then, it iterates through the bits of the BCD string in groups of 4, using the substr function to extract each group. For each group, it creates a bitset object with the 4 bits, and converts it to decimal using the to_ulong function. It then multiplies the decimal value by the appropriate place value and adds it to the num variable. Finally, the function returns num.

C++




#include <iostream>
#include <bitset>
using namespace std;
 
// Function to convert BCD to Decimal
int bcdToDecimal(string s)
{
    int len = s.length(), num = 0;
     
    // Iterating through the bits
    for (int i = 0; i < len; i+=4) {
 
        // Extracting the 4 bits from the BCD string
        bitset<4> bcd_bits(s.substr(i, 4));
 
        // Converting the 4 bits to decimal
        int dec = bcd_bits.to_ulong();
 
        // Multiplying by the place value and adding to the result
        num = num*10 + dec;
    }
 
    return num;
}
 
// Driver Code
int main()
{
    string s = "100000101000";
 
    // Function Call
    cout << bcdToDecimal(s);
 
    return 0;
}


Java




public class BCDToDecimal {
    // Function to convert a binary-coded decimal (BCD)
  // string to its decimal representation
    public static int bcdToDecimal(String s) {
        int len = s.length();
        int num = 0;
 
        // Iterate through the BCD string, reading 4 bits at a time
        for (int i = 0; i < len; i += 4) {
            // Extract the 4-bit BCD representation
            String bcdBits = s.substring(i, i + 4);
 
            // Convert the BCD representation to its decimal value
          // using Integer.parseInt
            int dec = Integer.parseInt(bcdBits, 2);
 
            // Append the decimal value to the result number
            num = num * 10 + dec;
        }
 
        // Return the final decimal number
        return num;
    }
 
    public static void main(String[] args) {
        String s = "100000101000";
        System.out.println(bcdToDecimal(s));
    }
}


Python3




def bcd_to_decimal(s):
    num = 0
    len_s = len(s)
     
    for i in range(0, len_s, 4):
        # Extracting the 4 bits from the BCD string
        bcd_bits = s[i:i+4]
 
        # Converting the 4 bits to decimal
        dec = int(bcd_bits, 2)
 
        # Multiplying by the place value and adding to the result
        num = num * 10 + dec
 
    return num
 
# Driver code
s = "100000101000"
 
# Function call
print(bcd_to_decimal(s))


C#




using System;
 
public class GFG
{
    // Function to convert BCD to Decimal
    public static int BcdToDecimal(string s)
    {
        int len = s.Length;
        int num = 0;
         
        // Iterating through the bits
        for (int i = 0; i < len; i += 4)
        {
            // Extracting the 4 bits from the BCD string
            string bcdSubstring = s.Substring(i, 4);
             
            // Parsing the 4 bits as a binary number
            int dec = Convert.ToInt32(bcdSubstring, 2);
             
            // Multiplying by the place value and adding to the result
            num = num * 10 + dec;
        }
 
        return num;
    }
 
    // Driver Code
    public static void Main()
    {
        string s = "100000101000";
 
        // Function Call
        Console.WriteLine(BcdToDecimal(s));
    }
}


Javascript




function bcdToDecimal(s) {
    const len = s.length;
    let num = 0;
 
    // Iterating through the bits
    for (let i = 0; i < len; i += 4) {
        // Extracting the 4 bits from the BCD string
        const bcdSubstring = s.substring(i, i + 4);
 
        // Parsing the 4 bits as a binary number
        const dec = parseInt(bcdSubstring, 2);
 
        // Multiplying by the place value and adding to the result
        num = num * 10 + dec;
    }
 
    return num;
}
 
// Driver Code
const s = "100000101000";
 
// Function Call
console.log(bcdToDecimal(s));


Output

828








The time complexity of this algorithm is O(n), where n is the length of the BCD string. This is because the algorithm must iterate through each bit of the string once. 

The auxiliary space complexity is also O(n), because the bitset objects created in each iteration require O(1) space, but there are n/4 iterations.
 



Last Updated : 20 Sep, 2023
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