How to add two Hexadecimal numbers?

Given two numeric Hexadecimal numbers str1 and str2, the task is to add the two hexadecimal numbers. 

Hexadecimal Number system, often shortened to “hex”, is a number system made up from 16 symbols. it uses 10 symbols from decimal number system which are represented by 0-9 and six extra symbols A – F which represent decimal 10 – 15.

Examples:

Input: str1 = “01B”, str2 = “378”
Output: 393
Explanation:
B (11 in decimal) + 8 =19 (13 in hex), hence addition bit = 3, carry = 1
1 + 7 + 1 (carry) = 9, hence addition bit = 9, carry = 0
0 + 3 + 0 (carry) = 3, hence addition bit = 3, carry = 0
01B + 378 = 393

Input: str1 = “AD”, str2 = “1B”
Output: C8
Explanation:
D(13 in Dec) + B(11 in Dec) = 24(18 in hex), hence addition bit = 8, carry = 1
A(10 in Dec) + 1 + 1 (carry)= 12 (C in hex), addition bit = C carry = 0
AD + 1B = C8  

Approaches:

• Using a map template to find and store the values.
• Using inbuilt functions to find the sum.

Method 1: Using maps

The idea is to use a map template to store the mapped values that are hexadecimal to decimal and decimal to hexadecimal.

1. Iterate until one of the given string reaches its length.
2. Start with carrying zero and add both numbers(with the carry) from the end and update carry in each addition.
3. Perform the same operation on the remaining length of the other string (if both the strings have different lengths).
4. Return the value that has been added.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Map for converting hexadecimal
// values to decimal
map<char, int> hex_value_of_dec(void)
{
    // Map the values to decimal values
    map<char, int> m{ { '0', 0 }, { '1', 1 },
                      { '2', 2 }, { '3', 3 },
                      { '4', 4 }, { '5', 5 },
                      { '6', 6 }, { '7', 7 },
                      { '8', 8 }, { '9', 9 },
                      { 'A', 10 }, { 'B', 11 },
                      { 'C', 12 }, { 'D', 13 },
                      { 'E', 14 }, { 'F', 15 } };

    return m;
}

// Map for converting decimal values
// to hexadecimal
map<int, char> dec_value_of_hex(void)
{
    // Map the values to the
    // hexadecimal values
    map<int, char> m{ { 0, '0' }, { 1, '1' },
                      { 2, '2' }, { 3, '3' },
                      { 4, '4' }, { 5, '5' },
                      { 6, '6' }, { 7, '7' },
                      { 8, '8' }, { 9, '9' },
                      { 10, 'A' }, { 11, 'B' },
                      { 12, 'C' }, { 13, 'D' },
                      { 14, 'E' }, { 15, 'F' } };

    return m;
}

// Function to add the two hexadecimal numbers
string Add_Hex(string a, string b)
{
    map<char, int> m = hex_value_of_dec();
    map<int, char> k = dec_value_of_hex();

    // Check if length of string first is
    // greater than or equal to string second
    if (a.length() < b.length())
        swap(a, b);

    // Store length of both strings
    int l1 = a.length(), l2 = b.length();

    string ans = "";

    // Initialize carry as zero
    int carry = 0, i, j;

    // Traverse till second string
    // get traversal completely
    for (i = l1 - 1, j = l2 - 1;
         j >= 0; i--, j--) {

        // Decimal value of element at a[i]
        // Decimal value of element at b[i]
        int sum = m[a[i]] + m[b[j]] + carry;

        // Hexadecimal value of sum%16
        // to get addition bit
        int addition_bit = k[sum % 16];

        // Add addition_bit to answer
        ans.push_back(addition_bit);

        // Update carry
        carry = sum / 16;
    }

    // Traverse remaining element
    // of string a
    while (i >= 0) {

        // Decimal value of element
        // at a[i]
        int sum = m[a[i]] + carry;

        // Hexadecimal value of sum%16
        // to get addition bit
        int addition_bit = k[sum % 16];

        // Add addition_bit to answer
        ans.push_back(addition_bit);

        // Update carry
        carry = sum / 16;
        i--;
    }

    // Check if still carry remains
    if (carry) {
        ans.push_back(k[carry]);
    }

    // Reverse the final string
    // for desired output
    reverse(ans.begin(), ans.end());

    // Return the answer
    return ans;
}

// Driver Code
int main(void)
{
    // Initialize the hexadecimal values
    string str1 = "1B", str2 = "AD";

    // Function call
    cout << Add_Hex(str1, str2) << endl;
}

Output

C8

Time Complexity: O(max(N, M)), where the length of the string first and second is N and M.
Auxiliary Space: O(max(N, M)), where the length of the string first and second is N and M.

Method 2: Using inbuilt functions

1. In python, there are inbuilt functions like hex() to convert binary numbers to hexadecimal numbers. 
2. To add two hexadecimal values in python we will first convert them into decimal values then add them and then finally again convert them to a hexadecimal value. 
3. To convert the numbers make use of the hex() function. 
4. The hex() function is one of the built-in functions in Python3, which is used to convert an integer number into its corresponding hexadecimal form.
5.  Use the int() function to convert the number to decimal form. The int() function in Python and Python3 converts a number in the given base to decimal.

Below is the implementation of the above approach:

Example 1:

Python3

# Program to add two hexadecimal numbers.

# Driver code
# Declaring the variables
str1 = "1B"
str2 = "AD"

# Calculating hexadecimal value using function
sum = hex(int(str1, 16) + int(str2, 16))

# Printing result
print(sum[2:])

Output

c8

Example 2:

Python3

# Python program to add two hexadecimal numbers.

# Driver code
if __name__ == "__main__" :

    # Declaring the variables
    a = "01C"
    b = "378"

    # Calculating hexadecimal sum by using hex() and int()
    hexadecimal_sum = lambda a,b : hex(int(a, 16) + int(b, 16))

    # calling hexadecimal_sum lambda function
    print(hexadecimal_sum(a,b)[2:])

    # This code is contributed by AnkThon

Output

394

Time complexity: O(1)
Auxiliary space: O(1)