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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