Smallest and Largest sum of two n-digit numbers

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Given an integer N ? 1, the task is to find the smallest and the largest sum of two N digit numbers.
Examples:

Input: N = 1
Output:
Largest = 18
Smallest = 0
Largest 1-digit number is 9 and 9 + 9 = 18
Smallest 1-digit number is 0 and 0 + 0 = 0
Input: N = 2
Output:
Largest = 198
Smallest = 20

Approach:

For largest: The answer will be 2 * (10^N – 1) because the series of sum of two n digit numbers will go on like 2 * 9, 2 * 99, 2 * 999, …
For smallest:
- If N = 1 then the answer will be 0.
- If N > 1 then the answer will be 2 * (10^{N – 1}) because the series of sum of two n digit numbers will go on like 0, 20, 200, 2000, …

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the smallest sum
// of 2 n-digit numbers
int smallestSum(int n)
{
    if (n == 1)
        return 0;
    return (2 * pow(10, n - 1));
}
 
// Function to return the largest sum
// of 2 n-digit numbers
int largestSum(int n)
{
    return (2 * (pow(10, n) - 1));
}
 
// Driver code
int main()
{
    int n = 4;
    cout << "Largest = " << largestSum(n) << endl;
    cout << "Smallest = " << smallestSum(n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG {
 
    // Function to return the smallest sum
    // of 2 n-digit numbers
    static int smallestSum(int n)
    {
        if (n == 1)
            return 0;
        return (2 * (int)Math.pow(10, n - 1));
    }
 
    // Function to return the largest sum
    // of 2 n-digit numbers
    static int largestSum(int n)
    {
        return (2 * ((int)Math.pow(10, n) - 1));
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 4;
        System.out.println("Largest = " + largestSum(n));
        System.out.print("Smallest = " + smallestSum(n));
    }
}

Python3

# Python3 implementation of the approach
 
# Function to return the smallest sum 
# of 2 n-digit numbers 
def smallestSum(n):
  
    if (n == 1):
        return 0
    return (2 * pow(10, n - 1))
 
# Function to return the largest sum 
# of 2 n-digit numbers 
def largestSum(n):
    return (2 * (pow(10, n) - 1))
 
# Driver code
n = 4
print("Largest = ", largestSum(n))
print("Smallest = ", smallestSum(n))

C#

// C# implementation of the approach
using System;
class GFG {
 
    // Function to return the smallest sum
    // of 2 n-digit numbers
    static int smallestSum(int n)
    {
        if (n == 1)
            return 0;
        return (2 * (int)Math.Pow(10, n - 1));
    }
 
    // Function to return the largest sum
    // of 2 n-digit numbers
    static int largestSum(int n)
    {
        return (2 * ((int)Math.Pow(10, n) - 1));
    }
 
    // Driver code
    public static void Main()
    {
        int n = 4;
        Console.WriteLine("Largest = " + largestSum(n));
        Console.Write("Smallest = " + smallestSum(n));
    }
}

PHP

<?php
// PHP implementation of the approach
 
// Function to return the smallest sum 
// of 2 n-digit numbers 
function smallestSum($n)
{
    if ($n == 1)
        return 0;
    return (2 * pow(10, $n - 1));
}
  
// Function to return the largest sum
// of 2 n-digit numbers
function largestSum($n)
{
    return 2 * ( pow(10, $n) - 1 );
}
 
// Driver code
$n = 4;
echo "Largest = " . largestSum($n) . "\n";
echo "Smallest = " . smallestSum($n);
?>

Javascript

<script>
 
// Javascript implementation of the approach
 
// Function to return the smallest sum
// of 2 n-digit numbers
function smallestSum(n)
{
    if (n == 1)
        return 0;
    return (2 * Math.pow(10, n - 1));
}
 
// Function to return the largest sum
// of 2 n-digit numbers
function largestSum(n)
{
    return (2 * (Math.pow(10, n) - 1));
}
 
// Driver code
var n = 4;
document.write("Largest = " + largestSum(n) + "<br>");
document.write("Smallest = " + smallestSum(n));
 
// This code is contributed by noob2000.
</script>

Output:

Largest = 19998
Smallest = 2000

Time Complexity: O(log n)

Auxiliary Space: O(1)

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