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 * (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, …
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 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
// 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
# 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 coden = 4print("Largest = ", largestSum(n))print("Smallest = ", smallestSum(n)) |
C#
// 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// 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, $n - 1));} // Function to return the largest sum// of 2 n-digit numbersfunction largestSum($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 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> |
Output:
Largest = 19998 Smallest = 2000
Time Complexity: O(log n)
Auxiliary Space: O(1)
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