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

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

Iterate until one of the given string reaches its length.
Start with carrying zero and add both numbers(with the carry) from the end and update carry in each addition.
Perform the same operation on the remaining length of the other string (if both the strings have different lengths).
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.

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My Personal Notes

C++

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