Smallest and Largest N-digit number starting and ending with N

Given an integer N, the task is to find the smallest and the largest N-digit numbers which start and ends with digit N.
Examples:

Input: N = 3 
Output:
Smallest Number = 303 
Largest Number = 393 
Explanation:
303 is the smallest 3 digit number starting and ending with 3. 
393 is the largest 3 digit number starting and ending with 3.
Input: N = 1 
Output:
Smallest Number = 1 
Largest Number = 1 
Explanation:
1 is both the smallest and the largest 1 digit number which starts and ends with 1. 

Approach:
We know that the largest and the smallest N-digit number is 9999…9, where 9 repeats N-times and 1000…. 0, where 0 repeats N-1 times respectively. 
Now to get the smallest and largest N-digit number starts and ends with N, we need to replace the first and the last digit of the smallest and the largest N-digit number by N. 
We have to take care of corner case i.e., when N = 1, here both the largest and the smallest number will be 1. 
Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find n digit
// largest number starting
// and ending with n
string findNumberL(int n)
{

    // Corner Case when n = 1
    if (n == 1)
        return "1";

    // Result will store the
    // n - 2*length(n) digit
    // largest number
    string result = "";

    // Find the number of
    // digits in number n
    int length = (int)floor(log10(n) + 1);

    // Append 9
    for(int i = 1; i <= n - (2 * length); i++)
    {
        result += '9';
    }

    // To make it largest n digit
    // number starting and ending
    // with n, we just need to
    // append n at start and end
    result = to_string(n) + result + to_string(n);

    // Return the largest number
    return result;
}

// Function to find n digit
// smallest number starting
// and ending with n
string findNumberS(int n)
{

    // Corner Case when n = 1
    if (n == 1)
        return "1";

    // Result will store the
    // n - 2*length(n) digit
    // smallest number
    string result = "";

    // Find the number of
    // digits in number n
    int length = (int)floor(log10(n) + 1);
    for (int i = 1; i <= n - (2 * length); i++)
    {
        result += '0';
    }

    // To make it smallest n digit
    // number starting and ending
    // with n, we just need to
    // append n at start and end
    result = to_string(n) + result + to_string(n);

    // Return the smallest number
    return result;
}

// Driver code
int main()
{

    // Given number
    int N = 3;

    // Function call
    cout << "Smallest Number = " << findNumberS(N) << endl;
    cout << "Largest Number = " << findNumberL(N);
    return 0;
}

// This code is contributed by divyeshrabadiya07

Java

// Java program for the above approach
import java.io.*;

class GFG {

    // Function to find n digit
    // largest number starting
    // and ending with n
    static String findNumberL(int n)
    {
        // Corner Case when n = 1
        if (n == 1)
            return "1";

        // Result will store the
        // n - 2*length(n) digit
        // largest number
        String result = "";

        // Find the number of
        // digits in number n
        int length
            = (int)Math.floor(
                Math.log10(n) + 1);

        // Append 9
        for (int i = 1;
             i <= n - (2 * length); i++) {
            result += '9';
        }

        // To make it largest n digit
        // number starting and ending
        // with n, we just need to
        // append n at start and end
        result = Integer.toString(n)
                 + result
                 + Integer.toString(n);

        // Return the largest number
        return result;
    }

    // Function to find n digit
    // smallest number starting
    // and ending with n
    static String findNumberS(int n)
    {

        // Corner Case when n = 1
        if (n == 1)
            return "1";

        // Result will store the
        // n - 2*length(n) digit
        // smallest number
        String result = "";

        // Find the number of
        // digits in number n
        int length
            = (int)Math.floor(
                Math.log10(n) + 1);
        for (int i = 1; i <= n - (2 * length); i++) {
            result += '0';
        }

        // To make it smallest n digit
        // number starting and ending
        // with n, we just need to
        // append n at start and end
        result = Integer.toString(n)
                 + result
                 + Integer.toString(n);

        // Return the smallest number
        return result;
    }

    // Driver Code
    public static void main(String[] args)
    {
        // Given Number
        int N = 3;

        // Function Call
        System.out.println(
            "Smallest Number = "
            + findNumberS(N));
        System.out.print(
            "Largest Number = "
            + findNumberL(N));
    }
}

Python3

# Python3 program for the
# above approach
import math

# Function to find n digit
# largest number starting
#and ending with n
def findNumberL(n):

    # Corner Case when n = 1
    if (n == 1):
        return "1"

    # Result will store the
    # n - 2*length(n) digit
    # largest number
    result = ""

    # Find the number of
    # digits in number n
    length  = math.floor(math.log10(n) + 1)

    # Append 9
    for i in range(1, n - (2 *
                   length) + 1):
        result += '9'

    # To make it largest n digit
    # number starting and ending
    # with n, we just need to
    # append n at start and end
    result = (str(n) + result +
              str(n))

    # Return the largest number
    return result

# Function to find n digit
# smallest number starting
# and ending with n
def findNumberS(n):

    # Corner Case when n = 1
    if (n == 1):
            return "1"

    # Result will store the
    # n - 2*length(n) digit
    # smallest number
    result = ""

    # Find the number of
    # digits in number n
    length = math.floor(math.log10(n) + 1)

    for i in range(1, n -
                   (2 * length) + 1):
        result += '0'

    # To make it smallest n digit
    # number starting and ending
    # with n, we just need to
    # append n at start and end
    result = (str(n) + result +
              str(n))

    # Return the smallest number
    return result

# Driver Code
if __name__ == "__main__":

    # Given Number
    N = 3

    # Function Call
    print("Smallest Number = " + findNumberS(N))
    print("Largest Number = "+ findNumberL(N))

# This code is contributed by Chitranayal

C#

// C# program for the above approach
using System;

class GFG{

// Function to find n digit
// largest number starting
// and ending with n
static String findNumberL(int n)
{

    // Corner Case when n = 1
    if (n == 1)
        return "1";

    // Result will store the
    // n - 2*length(n) digit
    // largest number
    String result = "";

    // Find the number of
    // digits in number n
    int length = (int)Math.Floor(
                      Math.Log10(n) + 1);

    // Append 9
    for(int i = 1;
            i <= n - (2 * length); i++)
    {
        result += '9';
    }

    // To make it largest n digit
    // number starting and ending
    // with n, we just need to
    // append n at start and end
    result = n.ToString() + result +
             n.ToString();

    // Return the largest number
    return result;
}

// Function to find n digit
// smallest number starting
// and ending with n
static String findNumberS(int n)
{

    // Corner Case when n = 1
    if (n == 1)
        return "1";

    // Result will store the
    // n - 2*length(n) digit
    // smallest number
    String result = "";

    // Find the number of
    // digits in number n
    int length = (int)Math.Floor(
                      Math.Log10(n) + 1);
    for (int i = 1;
             i <= n - (2 * length); i++)
    {
        result += '0';
    }

    // To make it smallest n digit
    // number starting and ending
    // with n, we just need to
    // append n at start and end
    result = n.ToString() + result +
             n.ToString();

    // Return the smallest number
    return result;
}

// Driver Code
public static void Main(String[] args)
{

    // Given number
    int N = 3;

    // Function call
    Console.WriteLine("Smallest Number = " +
                       findNumberS(N));
    Console.Write("Largest Number = " +
                   findNumberL(N));
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>

// JavaScript program for the above approach

    // Function to find n digit
    // largest number starting
    // and ending with n
    function findNumberL(n)
    {
        // Corner Case when n = 1
        if (n == 1)
            return "1";

        // Result will store the
        // n - 2*length(n) digit
        // largest number
        let result = "";

        // Find the number of
        // digits in number n
        let length
            = Math.floor(
                Math.log10(n) + 1);

        // Append 9
        for (let i = 1;
             i <= n - (2 * length); i++) {
            result += '9';
        }

        // To make it largest n digit
        // number starting and ending
        // with n, we just need to
        // append n at start and end
        result = n.toString()
                 + result
                 + n.toString();

        // Return the largest number
        return result;
    }

    // Function to find n digit
    // smallest number starting
    // and ending with n
    function findNumberS(n)
    {

        // Corner Case when n = 1
        if (n == 1)
            return "1";

        // Result will store the
        // n - 2*length(n) digit
        // smallest number
        let result = "";

        // Find the number of
        // digits in number n
        let length
            = Math.floor(
                Math.log10(n) + 1);
        for (let i = 1; i <= n - (2 * length); i++) {
            result += '0';
        }

        // To make it smallest n digit
        // number starting and ending
        // with n, we just need to
        // append n at start and end
        result = n.toString()
                 + result
                 + n.toString();

        // Return the smallest number
        return result;
    }


// Driver Code

     // Given Number
        let N = 3;

        // Function Call
        document.write(
            "Smallest Number = "
            + findNumberS(N) + "<br/>");
        document.write(
            "Largest Number = "
            + findNumberL(N));

</script>

Output:

Smallest Number = 303
Largest Number = 393

Time Complexity: O(N)

Space Complexity: O(N)