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:
- 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 the sum in a variable num.
-
Reverse the number formed so far and return that number.
Below is the implementation of the above approach.
// 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 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 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# 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 |
<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> |
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.
#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;
} |
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));
}
} |
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))
|
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));
}
} |
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)); |
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.