Program to Convert BCD number into Decimal number

Given a BCD (Binary Coded Decimal) number, the task is to convert the BCD number into it’s 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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

Iterate over all bits in given BCD numbers.
Divide the given BCD number into chunks of 4, and start computing its equivalent Decimal number.
Store this number formed in a variable named sum.
Start framing a number from the digits stored in sum in a variable num.
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 varaibles 
        // 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 varaibles 
       // 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 varaibles 
        # 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 varaibles 
        // 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 

Output:

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

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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