Smallest and Largest sum of two n-digit numbers - GeeksforGeeks

26 Mar, 2021
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 = 0Input: N = 2 Output: Largest = 198 Smallest = 20  
 
Approach:  
For largest: The answer will be 2 * (10N – 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 * (10N – 1) because the series of sum of two n digit numbers will go on like 0, 20, 200, 2000, ...
If N = 1 then the answer will be 0.
If N > 1 then the answer will be 2 * (10N – 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++
Java
Python3
C#
PHP
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Function to return the smallest sum// of 2 n-digit numbersint smallestSum(int n){    if (n == 1)        return 0;    return (2 * pow(10, n - 1));} // Function to return the largest sum// of 2 n-digit numbersint largestSum(int n){    return (2 * (pow(10, n) - 1));} // Driver codeint main(){    int n = 4;    cout << "Largest = " << largestSum(n) << endl;    cout << "Smallest = " << smallestSum(n);     return 0;}
// Java implementation of the approachclass 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 implementation of the approach # Function to return the smallest sum# of 2 n-digit numbersdef smallestSum(n):      if (n == 1):        return 0    return (2 * pow(10, n - 1)) # Function to return the largest sum# of 2 n-digit numbersdef largestSum(n):    return (2 * (pow(10, n) - 1)) # Driver coden = 4print("Largest = ", largestSum(n))print("Smallest = ", smallestSum(n))
// C# implementation of the approachusing 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 implementation of the approach // Function to return the smallest sum// of 2 n-digit numbersfunction smallestSum($n){    if ($n == 1)        return 0;    return (2 * pow(10, $n - 1));}  // Function to return the largest sum// of 2 n-digit numbersfunction largestSum($n){    return 2 * ( pow(10, $n) - 1 );} // Driver code$n = 4;echo "Largest = " . largestSum($n) . "\n";echo "Smallest = " . smallestSum($n);?>
<script> // Javascript implementation of the approach // Function to return the smallest sum// of 2 n-digit numbersfunction smallestSum(n){    if (n == 1)        return 0;    return (2 * Math.pow(10, n - 1));} // Function to return the largest sum// of 2 n-digit numbersfunction largestSum(n){    return (2 * (Math.pow(10, n) - 1));} // Driver codevar n = 4;document.write("Largest = " + largestSum(n) + "<br>");document.write("Smallest = " + smallestSum(n)); // This code is contributed by noob2000.</script>
Largest = 19998
Smallest = 2000
 
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