Find the remainder when First digit of a number is divided by its Last digit

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 remainder
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;

    cout << remainder << endl;
}

// Driver code
int 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 digit
import java.io.*;
class GFG
{

// Function to find the remainder
static 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 code
public 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 remainder
def 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 code
n = 5223

findRemainder(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 digit
using System;

class GFG
{

// Function to find the remainder
static 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 code
public 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 remainder
function 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:

1. Convert the input number to a string so that we can easily access its first and last digits.
2. Extract the first and last digits of the number by accessing the first and last characters of the string.
3. Convert the first digit to an integer using the ASCII code of ‘0’ and the subtraction operator.
4. Convert the last digit to an integer using the same method.
5. Find the remainder when the first digit is divided by the last digit using the modulo operator (%).
6. Output the remainder.

Implementation of the above approach:

C++

#include <iostream>
#include <string>

using namespace std;

// Function to find the remainder
void 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 code
int 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 code
n = 5223
find_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 remainder
function 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)