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++

   // CPP program to largest and smallest digit of a number 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to the largest and smallest digit of a number 
void Digits(int n) 
{ 
    int largest = 0; 
    int smallest = 9; 
  
    while (n) { 
        int r = n % 10; 
  
        // Find the largest digit 
        largest = max(r, largest); 
  
        // Find the smallest digit 
        smallest = min(r, smallest); 
  
        n = n / 10; 
    } 
    cout << largest << " " << smallest; 
} 
  
// Driver code 
int main() 
{ 
    int n = 2346; 
  
    // Function call 
    Digits(n); 
  
    return 0; 
} 
 
  

Java

   // Java program to largest and smallest digit of a number 
import java.util.*; 
import java.lang.*; 
import java.io.*; 
  
class Gfg 
{ 
      
// Function to the largest and smallest digit of a number 
static void Digits(int n) 
{ 
    int largest = 0; 
    int smallest = 9; 
  
    while(n != 0)  
    { 
        int r = n % 10; 
  
        // Find the largest digit 
        largest = Math.max(r, largest); 
  
        // Find the smallest digit 
        smallest = Math.min(r, smallest); 
  
        n = n / 10; 
    } 
    System.out.println(largest + " " + smallest); 
} 
  
// Driver code 
public static void main (String[] args) throws java.lang.Exception 
{ 
    int n = 2346; 
  
    // Function call 
    Digits(n); 
  
} 
} 
  
// This code is contributed by nidhiva 
 
  

Python3

   # Python3 program to largest and smallest digit of a number 
  
# Function to the largest and smallest digit of a number 
def Digits(n): 
    largest = 0
    smallest = 9
  
    while (n): 
        r = n % 10
  
        # Find the largest digit 
        largest = max(r, largest) 
  
        # Find the smallest digit 
        smallest = min(r, smallest) 
  
        n = n // 10
  
    print(largest,smallest) 
  
# Driver code 
  
n = 2346
  
# Function call 
Digits(n) 
  
# This code is contributed by mohit kumar 29

C#

   // C# program to largest and 
// smallest digit of a number 
using System; 
      
class GFG 
{ 
      
// Function to the largest and  
// smallest digit of a number 
static void Digits(int n) 
{ 
    int largest = 0; 
    int smallest = 9; 
  
    while(n != 0)  
    { 
        int r = n % 10; 
  
        // Find the largest digit 
        largest = Math.Max(r, largest); 
  
        // Find the smallest digit 
        smallest = Math.Min(r, smallest); 
  
        n = n / 10; 
    } 
    Console.WriteLine(largest + " " + smallest); 
} 
  
// Driver code 
public static void Main (String[] args) 
{ 
    int n = 2346; 
  
    // Function call 
    Digits(n); 
} 
} 
  
// This code is contributed by PrinciRaj1992  
 
  

Javascript

   <script> 
  
// Javascript program to largest and 
// smallest digit of a number 
  
// Function to the largest and smallest 
// digit of a number 
function Digits(n) 
{ 
    let largest = 0; 
    let smallest = 9; 
  
    while (n) { 
        let r = n % 10; 
  
        // Find the largest digit 
        largest = Math.max(r, largest); 
  
        // Find the smallest digit 
        smallest = Math.min(r, smallest); 
  
        n = parseInt(n / 10); 
    } 
    document.write(largest + " " + smallest); 
} 
  
// Driver code 
    let n = 2346; 
  
    // Function call 
    Digits(n); 
  
</script>
 
  

Output

6 2

Time Complexity: O(log(n)), where n is the given number
Auxiliary Space: O(1)

Approach : Using str(),min(),max()

C++

   // C++ program to largest and smallest digit of a number 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to the largest and smallest digit of a number 
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; 
} 
  
// This code is contributed by phasing17
 
  

Java

   // Java program to largest and smallest digit of a number 
import java.util.stream.Collectors.*; 
import java.util.stream.*; 
import java.util.*; 
  
class GFG 
{ 
    
  // Function to the largest and smallest digit of a number 
  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); 
  } 
} 
  
// This code is contributed by phasing17
 
  

Python3

   # Python3 program to largest and smallest digit of a number 
  
# Function to the largest and smallest digit of a number 
n=2346
s=str(n) 
print(max(s),end=" ") 
print(min(s))
 
  

C#

   // C# program to largest and smallest digit of a number 
using System; 
using System.Linq; 
using System.Collections.Generic; 
  
class GFG 
{ 
    // Function to the largest and smallest digit of a number 
    public static void Main(string[] args) 
    { 
        int n = 2346; 
        char[] s = Convert.ToString(n).ToCharArray(); 
          
        Console.WriteLine(s.Max() + " " + s.Min()); 
    } 
} 
  
// This code is contributed by phasing17
 
  

Javascript

   // JS program to largest and smallest digit of a number 
  
// Function to the largest and smallest digit of a number 
let n=2346 
let s = ("" + n).split("") 
  
console.log(s[s.length - 1]) 
console.log(s[0]) 
  
// This code is contributed by phasing17

Output

6 2

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; 
  
    // Convert string to a vector of integers 
    for (char c : n) { 
        digits.push_back(c - '0'); 
    } 
  
    // Find the smallest and largest digits using the 
    // accumulate function 
    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); }); 
  
    // Print the smallest and largest digits 
    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<>(); 
  
        // Convert string to a list of integers 
        for (char c : n.toCharArray()) { 
            digits.add(Character.getNumericValue(c)); 
        } 
  
        // Find the smallest and largest digits using the accumulate function 
        int smallest = digits.stream().reduce(digits.get(0), Integer::min); 
        int largest = digits.stream().reduce(digits.get(0), Integer::max); 
  
        // Print the smallest and largest digits 
        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"; 
          
        // Convert string to a List of integers 
        var digits = n.Select(c => c - '0').ToList(); 
  
        // Find the smallest and largest digits using the 
        // Aggregate function 
        int smallest = digits.Aggregate((x, y) => Math.Min(x, y)); 
        int largest = digits.Aggregate((x, y) => Math.Max(x, y)); 
  
        // Print the smallest and largest digits 
        Console.WriteLine("Smallest digit: " + smallest); 
        Console.WriteLine("Largest digit: " + largest); 
    } 
} 
 
  

Javascript

   // Define the input string 
const n = "2346"; 
const digits = []; 
  
// Convert string to an array of integers 
for (const c of n) { 
    digits.push(parseInt(c)); 
} 
  
// Find the smallest and largest digits using the reduce function 
const smallest = digits.reduce((x, y) => Math.min(x, y), digits[0]); 
const largest = digits.reduce((x, y) => Math.max(x, y), digits[0]); 
  
// Print the smallest and largest digits 
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)

C++ Program for Smallest K digit number divisible by X

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Largest and smallest digit of a number

C++

Java

Python3

C#

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C++

Java

Python3

C#

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Approach#4: Using reduce

Algorithm

C++

Java

Python3

C#

Javascript

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