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.
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; } |
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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 |
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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 |
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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 |
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Output:
828
Time complexity: O(N)
Auxiliary Space complexity: O(1)
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