Largest and smallest digit of a number
Last Updated :
13 Dec, 2023
Given a number N. The task is to find the largest and the smallest digit of the number.
Examples :
Input : N = 2346
Output : 6 2
6 is the largest digit and 2 is smallest
Input : N = 5
Output : 5 5
Approach: An efficient approach is to find all digits in the given number and find the largest and the smallest digit.
C++
#include <bits/stdc++.h>
using namespace std;
void Digits(int n)
{
int largest = 0;
int smallest = 9;
while (n) {
int r = n % 10;
largest = max(r, largest);
smallest = min(r, smallest);
n = n / 10;
}
cout << largest << " " << smallest;
}
int main()
{
int n = 2346;
Digits(n);
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
import java.io.*;
class Gfg
{
static void Digits(int n)
{
int largest = 0;
int smallest = 9;
while(n != 0)
{
int r = n % 10;
largest = Math.max(r, largest);
smallest = Math.min(r, smallest);
n = n / 10;
}
System.out.println(largest + " " + smallest);
}
public static void main (String[] args) throws java.lang.Exception
{
int n = 2346;
Digits(n);
}
}
|
Python3
def Digits(n):
largest = 0
smallest = 9
while (n):
r = n % 10
largest = max(r, largest)
smallest = min(r, smallest)
n = n // 10
print(largest,smallest)
n = 2346
Digits(n)
|
C#
using System;
class GFG
{
static void Digits(int n)
{
int largest = 0;
int smallest = 9;
while(n != 0)
{
int r = n % 10;
largest = Math.Max(r, largest);
smallest = Math.Min(r, smallest);
n = n / 10;
}
Console.WriteLine(largest + " " + smallest);
}
public static void Main (String[] args)
{
int n = 2346;
Digits(n);
}
}
|
Javascript
<script>
function Digits(n)
{
let largest = 0;
let smallest = 9;
while (n) {
let r = n % 10;
largest = Math.max(r, largest);
smallest = Math.min(r, smallest);
n = parseInt(n / 10);
}
document.write(largest + " " + smallest);
}
let n = 2346;
Digits(n);
</script>
|
Time Complexity: O(log(n)), where n is the given number
Auxiliary Space: O(1)
Approach : Using str(),min(),max()
C++
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n = 2346;
string s = to_string(n);
cout << *(max_element(s.begin(), s.end())) << " ";
cout << *(min_element(s.begin(), s.end())) << endl;
}
|
Java
import java.util.stream.Collectors.*;
import java.util.stream.*;
import java.util.*;
class GFG
{
public static void main(String[] args)
{
int n = 2346;
String s1 = String.valueOf(n);
List<Character> s = s1.chars().mapToObj(c -> (char) c).collect(Collectors.toList());
Comparator<Character> comparator = Comparator.comparing( Character::valueOf );
Character mins = s.stream().min(comparator).get();
Character maxs = s.stream().max(comparator).get();
System.out.println(maxs + " " + mins);
}
}
|
Python3
n=2346
s=str(n)
print(max(s),end=" ")
print(min(s))
|
C#
using System;
using System.Linq;
using System.Collections.Generic;
class GFG
{
public static void Main(string[] args)
{
int n = 2346;
char[] s = Convert.ToString(n).ToCharArray();
Console.WriteLine(s.Max() + " " + s.Min());
}
}
|
Javascript
let n=2346
let s = ("" + n).split("")
console.log(s[s.length - 1])
console.log(s[0])
|
Time Complexity: O(1)
Auxiliary Space: O(1)
Approach#4: Using reduce
Convert the input number to a list of integers (digits). Use the reduce function to apply a lambda function that compares two digits at a time and returns the smallest or largest.
Algorithm
1. Convert the input number to a list of integers using map and list.
2. Use reduce function from functools module to find the smallest and largest digit.
3. In the lambda function passed to reduce, the two arguments x and y represent two adjacent digits in the list. The lambda function compares these two digits and returns the smallest or largest based on the condition.
4. The reduce function applies this lambda function to the entire list of digits, returning the smallest and largest digit.
C++
#include <algorithm>
#include <iostream>
#include <numeric> // Include this header for the accumulate function
#include <vector>
int main()
{
std::string n = "2346";
std::vector<int> digits;
for (char c : n) {
digits.push_back(c - '0');
}
int smallest = std::accumulate(
digits.begin(), digits.end(), digits[0],
[](int x, int y) { return std::min(x, y); });
int largest = std::accumulate(
digits.begin(), digits.end(), digits[0],
[](int x, int y) { return std::max(x, y); });
std::cout << "Smallest digit: " << smallest
<< std::endl;
std::cout << "Largest digit: " << largest << std::endl;
return 0;
}
|
Java
import java.util.ArrayList;
import java.util.List;
public class Main {
public static void main(String[] args) {
String n = "2346";
List<Integer> digits = new ArrayList<>();
for (char c : n.toCharArray()) {
digits.add(Character.getNumericValue(c));
}
int smallest = digits.stream().reduce(digits.get(0), Integer::min);
int largest = digits.stream().reduce(digits.get(0), Integer::max);
System.out.println("Smallest digit: " + smallest);
System.out.println("Largest digit: " + largest);
}
}
|
Python3
from functools import reduce
n = '2346'
digits = list(map(int, n))
smallest = reduce(lambda x, y: x if x < y else y, digits)
largest = reduce(lambda x, y: x if x > y else y, digits)
print("Smallest digit:", smallest)
print("Largest digit:", largest)
|
C#
using System;
using System.Linq;
class GFG
{
static void Main()
{
string n = "2346";
var digits = n.Select(c => c - '0').ToList();
int smallest = digits.Aggregate((x, y) => Math.Min(x, y));
int largest = digits.Aggregate((x, y) => Math.Max(x, y));
Console.WriteLine("Smallest digit: " + smallest);
Console.WriteLine("Largest digit: " + largest);
}
}
|
Javascript
const n = "2346";
const digits = [];
for (const c of n) {
digits.push(parseInt(c));
}
const smallest = digits.reduce((x, y) => Math.min(x, y), digits[0]);
const largest = digits.reduce((x, y) => Math.max(x, y), digits[0]);
console.log("Smallest digit: " + smallest);
console.log("Largest digit: " + largest);
|
Output
Smallest digit: 2
Largest digit: 6
Time complexity: O(n)
Auxiliary Space: O(n)
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