Given a number N, find the remainder when the first digit of N is divided by its last digit.
Examples:
Input: N = 1234 Output: 1 First digit = 1 Last digit = 4 Remainder = 1 % 4 = 1 Input: N = 5223 Output: 2 First digit = 5 Last digit = 3 Remainder = 5 % 3 = 2
Approach: Find the first digit and the last digit of the number. Find then the remainder when the first digit is divided by the last digit.
Below is the implementation of the above approach:
C++
// C++ program to find the remainder// when the First digit of a number// is divided by its Last digit#include <bits/stdc++.h>using namespace std;// Function to find the remaindervoid findRemainder(int n){ // Get the last digit int l = n % 10; // Get the first digit while (n >= 10) n /= 10; int f = n; // Compute the remainder int remainder = f % l; cout << remainder << endl;}// Driver codeint main(){ int n = 5223; findRemainder(n); return 0;} |
Java
// Java program to find the remainder// when the First digit of a number// is divided by its Last digitimport java.io.*;class GFG{ // Function to find the remainderstatic void findRemainder(int n){ // Get the last digit int l = n % 10; // Get the first digit while (n >= 10) n /= 10; int f = n; // Compute the remainder int remainder = f % l; System.out.println(remainder);}// Driver codepublic static void main(String[] args){ int n = 5223; findRemainder(n);}}// This code is contributed by Code_Mech |
Python3
# Python3 program to find the remainder# when the First digit of a number# is divided by its Last digit# Function to find the remainderdef findRemainder(n): # Get the last digit l = n % 10 # Get the first digit while (n >= 10): n //= 10 f = n # Compute the remainder remainder = f % l print(remainder)# Driver coden = 5223findRemainder(n)# This code is contributed by Mohit Kumar |
C#
// C# program to find the remainder// when the First digit of a number// is divided by its Last digitusing System;class GFG{ // Function to find the remainderstatic void findRemainder(int n){ // Get the last digit int l = n % 10; // Get the first digit while (n >= 10) n /= 10; int f = n; // Compute the remainder int remainder = f % l; Console.WriteLine(remainder);}// Driver codepublic static void Main(){ int n = 5223; findRemainder(n);}}// This code is contributed by Code_Mech |
Javascript
<script> // Javascript program to find the remainder// when the First digit of a number// is divided by its Last digit// Function to find the remainderfunction findRemainder( n){ // Get the last digit let l = n % 10; // Get the first digit while (n >= 10) n /= 10; let f = n; // Compute the remainder let remainder = f % l; document.write(Math.floor(remainder));}// Driver code let n = 5223; findRemainder(n);// This code is contributed by mohan pavan</script> |
Output:
2
Time Complexity: O(L ) where L is length of number in decimal representation
Auxiliary Space: O(1)
Approach:
- Convert the input number to a string so that we can easily access its first and last digits.
- Extract the first and last digits of the number by accessing the first and last characters of the string.
- Convert the first digit to an integer using the ASCII code of ‘0’ and the subtraction operator.
- Convert the last digit to an integer using the same method.
- Find the remainder when the first digit is divided by the last digit using the modulo operator (%).
- Output the remainder.
Implementation of the above approach:
C++
#include <iostream>#include <string>using namespace std;// Function to find the remaindervoid findRemainder(int n){ string s = to_string(n); char first = s[0]; char last = s[s.length()-1]; int remainder = (first - '0') % (last - '0'); cout << remainder << endl;} // Driver codeint main(){ int n = 5223; findRemainder(n); return 0;} |
Java
import java.util.*;public class Main { // Function to find the remainder static void findRemainder(int n) { String s = Integer.toString(n); char first = s.charAt(0); char last = s.charAt(s.length()-1); int remainder = (first - '0') % (last - '0'); System.out.println(remainder); } // Driver code public static void main(String[] args) { int n = 5223; findRemainder(n); }}// This code is contributed by Prajwal Kandekar |
Python3
def find_remainder(n): s = str(n) first = s[0] last = s[-1] remainder = int(first) % int(last) print(remainder)# Driver coden = 5223find_remainder(n)# This code is contributed by Prajwal Kandekar |
C#
using System;namespace RemainderFinder {class Program { // Function to find the remainder static void FindRemainder(int n) { string s = n.ToString(); char first = s[0]; char last = s[s.Length - 1]; int remainder = (first - '0') % (last - '0'); Console.WriteLine(remainder); } // Driver code static void Main(string[] args) { int n = 5223; FindRemainder(n); Console.ReadLine(); }}}// This code is contributed by sarojmcy2e |
Javascript
// Function to find the remainderfunction findRemainder(n) { let s = n.toString(); let first = s.charAt(0); let last = s.charAt(s.length-1); let remainder = parseInt(first) % parseInt(last); console.log(remainder);}let n = 5223;findRemainder(n);// This code is contributed by Prajwal Kandekar |
Output
2
Time Complexity: O(1)
Space Complexity: O(1)
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