Given two hexadecimal numbers N and K, the task is to find N modulo K.
Examples:
Input: N = 3E8, K = 13
Output: C
Explanation:
Decimal representation of N( = 3E8) is 1000
Decimal representation of K( = 13) is 19
Decimal representation of (N % K) = 1000 % 19 = 12 ( = C).
Therefore, the required output is C.Input: N = 2A3, K = 1A
Output: 19
Approach: Follow the steps below to solve the problem:
- Convert the hexadecimal numbers, N and K into their equivalent decimal numbers say, X and Y respectively.
- Convert the decimal number, (X % Y) into its equivalent hexadecimal numbers say, res
- Finally, print the value of res.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate modulus of// two Hexadecimal numbersvoid hexaModK(string s, string k){ // Store all possible // hexadecimal digits map<char, int> mp; // Iterate over the range ['0', '9'] for(char i = 1; i <= 9; i++) { mp[i + '0'] = i; } mp['A'] = 10; mp['B'] = 11; mp['C'] = 12; mp['D'] = 13; mp['E'] = 14; mp['F'] = 15; // Convert given string to long long m = stoi(k, 0, 16); // Base to get 16 power long base = 1; // Store N % K long ans = 0; // Iterate over the digits of N for(int i = s.length() - 1; i >= 0; i--) { // Stores i-th digit of N long n = mp[s[i]] % m; // Update ans ans = (ans + (base % m * n % m) % m) % m; // Update base base = (base % m * 16 % m) % m; } // Print the answer converting // into hexadecimal stringstream ss; ss << hex << ans; string su = ss.str(); transform(su.begin(), su.end(), su.begin(), ::toupper); cout << (su);}// Driver Codeint main() { // Given string N and K string n = "3E8"; string k = "13"; // Function Call hexaModK(n, k); return 0;}// This code is contributed by sallagondaavinashreddy7 |
Java
// Java program to implement // the above approach import java.util.*; public class Main { // Function to calculate modulus of // two Hexadecimal numbers static void hexaModK(String N, String k) { // Store all possible // hexadecimal digits HashMap<Character, Integer> map = new HashMap<>(); // Iterate over the range ['0', '9'] for (char i = '0'; i <= '9'; i++) { map.put(i, i - '0'); } map.put('A', 10); map.put('B', 11); map.put('C', 12); map.put('D', 13); map.put('E', 14); map.put('F', 15); // Convert given string to long long m = Long.parseLong(k, 16); // Base to get 16 power long base = 1; // Store N % K long ans = 0; // Iterate over the digits of N for (int i = N.length() - 1; i >= 0; i--) { // Stores i-th digit of N long n = map.get(N.charAt(i)) % m; // Update ans ans = (ans + (base % m * n % m) % m) % m; // Update base base = (base % m * 16 % m) % m; } // Print the answer converting // into hexadecimal System.out.println( Long.toHexString(ans).toUpperCase()); } // Driver Code public static void main(String args[]) { // Given string N and K String n = "3E8"; String k = "13"; // Function Call hexaModK(n, k); } } |
Python3
# Python3 program to implement# the above approach# Function to calculate modulus of# two Hexadecimal numbersdef hexaModK(s, k) : # Store all possible # hexadecimal digits mp = {}; # Iterate over the range ['0', '9'] for i in range(1, 10) : mp[chr(i + ord('0'))] = i; mp['A'] = 10; mp['B'] = 11; mp['C'] = 12; mp['D'] = 13; mp['E'] = 14; mp['F'] = 15; # Convert given string to long m = int(k); # Base to get 16 power base = 1; # Store N % K ans = 0; # Iterate over the digits of N for i in range(len(s) - 1, -1, -1) : # Stores i-th digit of N n = mp[s[i]] % m; # Update ans ans = (ans + (base % m * n % m) % m) % m; # Update base base = (base % m * 16 % m) % m; # Print the answer converting # into hexadecimal ans = hex(int(ans))[-1].upper() print(ans)# Driver Codeif __name__ == "__main__" : # Given string N and K n = "3E8"; k = "13"; # Function Call hexaModK(n, k); # This code is contributed by AnkThon |
C#
// C# program to implement // the above approach using System;using System.Collections.Generic; class GFG{ // Function to calculate modulus of // two Hexadecimal numbers static void hexaModK(String N, String k) { // Store all possible // hexadecimal digits Dictionary<char, int> map = new Dictionary<char, int>(); // Iterate over the range ['0', '9'] for(char i = '0'; i <= '9'; i++) { map.Add(i, i - '0'); } map.Add('A', 10); map.Add('B', 11); map.Add('C', 12); map.Add('D', 13); map.Add('E', 14); map.Add('F', 15); // Convert given string to long long m = long.Parse(k); // Base to get 16 power long Base = 1; // Store N % K long ans = 0; // Iterate over the digits of N for(int i = N.Length - 1; i >= 0; i--) { // Stores i-th digit of N long n = map[N[i]] % m; // Update ans ans = (ans + (Base % m * n % m) % m) % m; // Update base Base = (Base % m * 16 % m) % m; } // Print the answer converting // into hexadecimal Console.WriteLine(ans.ToString("X")); } // Driver Code public static void Main(String []args) { // Given string N and K String n = "3E8"; String k = "13"; // Function Call hexaModK(n, k); } } // This code is contributed by Princi Singh |
Javascript
<script>// Javascript program to implement// the above approach// Function to calculate modulus of// two Hexadecimal numbersfunction hexaModK(s, k){ // Store all possible // hexadecimal digits var mp = new Map(); // Iterate over the range ['0', '9'] for(var i = 1; i <= 9; i++) { mp.set(String.fromCharCode( i + '0'.charCodeAt(0)), i); } mp.set('A', 10); mp.set('B', 11); mp.set('C', 12); mp.set('D', 13); mp.set('E', 14); mp.set('F', 15); // Convert given string to long var m = parseInt(k, 16); // Base to get 16 power var base = 1; // Store N % K var ans = 0; // Iterate over the digits of N for(var i = s.length - 1; i >= 0; i--) { // Stores i-th digit of N var n = mp.get(s[i]) % m; // Update ans ans = (ans + (base % m * n % m) % m) % m; // Update base base = (base % m * 16 % m) % m; } document.write(ans.toString(16).toUpperCase());}// Driver Code// Given string N and Kvar n = "3E8";var k = "13";// Function CallhexaModK(n, k);// This code is contributed by famously</script> |
Output:
C
Time Complexity: O(N)
Auxiliary Space: O(1)
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