Smallest and Largest N-digit perfect squares

Given an integer N, the task is to find the smallest and the largest N digit numbers which are also perfect squares.

Examples:

Input: N = 2
Output: 16 81
16 and 18 are the smallest and the largest 2-digit perfect squares.



Input: N = 3
Output: 100 961

Approach: For increasing values of N starting from N = 1, the series will go on like 9, 81, 961, 9801, ….. for the largest N-digit perfect square whose Nth term will be pow(ceil(sqrt(pow(10, N))) – 1, 2).
And 1, 16, 100, 1024, ….. for the smallest N-digit perfect square whose Nth term will be pow(ceil(sqrt(pow(10, N – 1))), 2).

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to print the largest and
// the smallest n-digit perfect squares
void nDigitPerfectSquares(int n)
{
  
    // Smallest n-digit perfect square
    cout << pow(ceil(sqrt(pow(10, n - 1))), 2) << " ";
  
    // Largest n-digit perfect square
    cout << pow(ceil(sqrt(pow(10, n))) - 1, 2);
}
  
// Driver code
int main()
{
    int n = 4;
    nDigitPerfectSquares(n);
  
    return 0;
}

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Java














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// Java implementation of the approach 


class GFG { 


  


    // Function to print the largest and 


    // the smallest n-digit perfect squares 


    static void nDigitPerfectSquares(int n) 


    { 


        // Smallest n-digit perfect square 


        int smallest = (int)Math.pow(Math.ceil(Math.sqrt(Math.pow(10, n - 1))), 2); 


        System.out.print(smallest + " "); 


  


        // Largest n-digit perfect square 


        int largest = (int)Math.pow(Math.ceil(Math.sqrt(Math.pow(10, n))) - 1, 2); 


        System.out.print(largest); 


    } 


  


    // Driver code 


    public static void main(String args[]) 


    { 


        int n = 4; 


        nDigitPerfectSquares(n); 


    } 


} 



















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Python3














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# Python3 implementation of the approach 


import math 


  


# Function to print the largest and 


# the smallest n-digit perfect squares 


def nDigitPerfectSquares(n): 


  


    # Smallest n-digit perfect square 


    print(pow(math.ceil(math.sqrt(pow(10, n - 1))), 2),  


                                            end = " "); 


  


    # Largest n-digit perfect square 


    print(pow(math.ceil(math.sqrt(pow(10, n))) - 1, 2)); 


  


# Driver code 


n = 4; 


nDigitPerfectSquares(n); 


  


# This code is contributed by mits 



















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C#














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// C# implementation of the approach 


using System; 


public class GFG { 


   


    // Function to print the largest and 


    // the smallest n-digit perfect squares 


    static void nDigitPerfectSquares(int n) 


    { 


        // Smallest n-digit perfect square 


        int smallest = (int)Math.Pow(Math.Ceiling(Math.Sqrt(Math.Pow(10, n - 1))), 2); 


        Console.Write(smallest + " "); 


   


        // Largest n-digit perfect square 


        int largest = (int)Math.Pow(Math.Ceiling(Math.Sqrt(Math.Pow(10, n))) - 1, 2); 


        Console.Write(largest); 


    } 


   


    // Driver code 


    public static void Main(String []args) 


    { 


        int n = 4; 


        nDigitPerfectSquares(n); 


    } 


} 


  


// This code has been contributed by 29AjayKumar 



















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PHP














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<?php 


// PHP implementation of the approach 


  


// Function to print the largest and 


// the smallest n-digit perfect squares 


function nDigitPerfectSquares($n) 


{ 


  


    // Smallest n-digit perfect square 


    echo pow(ceil(sqrt(pow(10, $n - 1))), 2), " "; 


  


    // Largest n-digit perfect square 


    echo pow(ceil(sqrt(pow(10, $n))) - 1, 2); 


} 


  


// Driver code 


$n = 4; 


nDigitPerfectSquares($n); 


  


// This code is contributed by jit_t 


?> 



















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Output:


1024 9801






          
          
          
          

            

    

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