Given a number to find the first and last digit of a number.
Examples:
Input : 12345
Output : First digit: 1
last digit : 5
Input : 98562
Output : First digit: 9
last digit : 2To find last digit of a number, we use modulo operator %. When modulo divided by 10 returns its last digit.
Suppose if n = 1234
then last Digit = n % 10 => 4
To find first digit of a number is little expensive than last digit. To find first digit of a number we divide the given number by 10 until number is greater than 10. At the end we are left with the first digit.
Approach 1 (With loop):
C++
#include <bits/stdc++.h>
using namespace std;
int firstDigit(int n)
{
while (n >= 10)
n /= 10;
return n;
}
int lastDigit(int n)
{
return (n % 10);
}
int main()
{
int n = 98562;
cout << firstDigit(n) << " "
<< lastDigit(n) << endl;
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
public class GfG{
public static int firstDigit(int n)
{
while (n >= 10)
n /= 10;
return n;
}
public static int lastDigit(int n)
{
return (n % 10);
}
public static void main(String argc[])
{
int n = 98562;
System.out.println(firstDigit(n) + " "
+ lastDigit(n));
}
}
|
Python3
def firstDigit(n) :
while n >= 10:
n = n / 10;
return int(n)
def lastDigit(n) :
return (n % 10)
n = 98562;
print(firstDigit(n), end = " ")
print(lastDigit(n))
|
C#
using System;
public class GfG{
public static int firstDigit(int n)
{
while (n >= 10)
n /= 10;
return n;
}
public static int lastDigit(int n)
{
return (n % 10);
}
public static void Main()
{
int n = 98562;
Console.WriteLine(firstDigit(n) + " "
+ lastDigit(n));
}
}
|
PHP
<?php
function firstDigit($n)
{
while ($n >= 10)
$n /= 10;
return (int)$n;
}
function lastDigit($n)
{
return ((int)$n % 10);
}
$n = 98562;
echo firstDigit($n) . " " .
lastDigit($n) . "\n";
|
Javascript
<script>
function firstDigit(n)
{
while (n >= 10)
n /= 10;
return Math.floor(n);
}
function lastDigit(n)
{
return Math.floor(n % 10);
}
let n = 98562;
document.write(firstDigit(n) + " "
+ lastDigit(n));
</script>
|
Time Complexity: O(log10n)
Auxiliary Space: O(1)
Approach 2 (Without loop)
C++
#include <bits/stdc++.h>
using namespace std;
int firstDigit(int n)
{
int digits = (int)log10(n);
n = (int)(n / pow(10, digits));
return n;
}
int lastDigit(int n)
{
return (n % 10);
}
int main()
{
int n = 98562;
cout << firstDigit(n) << " "
<< lastDigit(n) << endl;
return 0;
}
|
Java
import java.math.*;
class GFG {
static int firstDigit(int n)
{
int digits = (int)(Math.log10(n));
n = (int)(n / (int)(Math.pow(10, digits)));
return n;
}
static int lastDigit(int n)
{
return (n % 10);
}
public static void main(String args[])
{
int n = 98562;
System.out.println(firstDigit(n) +
" " + lastDigit(n));
}
}
|
Python3
import math
def firstDigit(n) :
digits = (int)(math.log10(n))
n = (int)(n / pow(10, digits))
return n;
def lastDigit(n) :
return (n % 10)
n = 98562;
print(firstDigit(n), end = " ")
print(lastDigit(n))
|
C#
using System;
class GFG {
static int firstDigit(int n)
{
int digits = (int)(Math.Log10(n));
n = (int)(n / (int)(Math.Pow(10, digits)));
return n;
}
static int lastDigit(int n)
{
return (n % 10);
}
public static void Main()
{
int n = 98562;
Console.WriteLine(firstDigit(n) +
" " + lastDigit(n));
}
}
|
PHP
<?php
function firstDigit($n)
{
$digits = (int)log10($n);
$n = (int)($n / pow(10, $digits));
return $n;
}
function lastDigit($n)
{
return ($n % 10);
}
$n = 98562;
echo firstDigit($n) , " ",
lastDigit($n), "\n";
?>
|
Javascript
<script>
function firstDigit(n)
{
let digits = Math.floor(Math.log(n)/Math.log(10))
n = Math.floor(n / Math.pow(10, digits))
return n;
}
function lastDigit(n){
return (n % 10)
}
let n = 98562;
document.write(firstDigit(n)," ")
document.write(lastDigit(n),"</br>")
</script>
|
Time Complexity: O(log(log10(n))
Auxiliary Space: O(1)
Important note: log10() is a mathematical function present in math.h header file. It returns log base 10 value of the passed parameter to log10() function.
Approach 3: Using string operation to_string
First converting number to string and then first char of string and last character of string – ‘0’ is our ans.
C++
#include <iostream>
using namespace std;
int main() {
int n = 34356;
string s = to_string(n);
int first_digit = s.front() - '0';
int last_digit = s.back() - '0';
cout<<"First digit of "<<n<<" is "<<first_digit<<endl;
cout<<"Last digit of "<<n<<" is "<<last_digit<<endl;
}
|
Java
import java.util.*;
public class Main {
public static void main(String[] args) {
int n = 34356;
String s = Integer.toString(n);
int first_digit = s.charAt(0) - '0';
int last_digit = s.charAt(s.length() - 1) - '0';
System.out.println("First digit of " + n + " is " + first_digit);
System.out.println("Last digit of " + n + " is " + last_digit);
}
}
|
Python3
n = 34356
s = str(n)
first_digit = int(s[0])
last_digit = int(s[-1])
print("First digit of", n, "is", first_digit)
print("Last digit of", n, "is", last_digit)
|
C#
using System;
public class GFG{
static public void Main (){
int n = 34356;
string s = n.ToString();
int first_digit = s[0] - '0';
int last_digit = s[s.Length - 1] - '0';
Console.WriteLine("First digit of {0} is {1}", n, first_digit);
Console.WriteLine("Last digit of {0} is {1}", n, last_digit);
}
}
|
Javascript
let n = 34356;
let s = n.toString();
let first_digit = parseInt(s.charAt(0));
let last_digit = parseInt(s.charAt(s.length-1));
console.log("First digit of " + n + " is " + first_digit);
console.log("Last digit of " + n + " is " + last_digit);
|
OutputFirst digit of 34356 is 3
Last digit of 34356 is 6
Time Complexity: O(n)
Auxiliary Space: O(n)