Smallest and Largest N-digit perfect squares

Difficulty Level : Medium
Last Updated : 26 Mar, 2021

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++

// 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;
}

Java

// 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);
    }
}

Python3

# 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

C#

// 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

PHP

<?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
?>

Javascript

<script>
 
// Javascript implementation of the approach
 
// Function to print the largest and
// the smallest n-digit perfect squares
function nDigitPerfectSquares(n)
{
 
    // Smallest n-digit perfect square
    document.write(Math.pow(Math.ceil(Math.sqrt(Math.pow(10, n - 1))), 2) + " ");
 
    // Largest n-digit perfect square
    document.write(Math.pow(Math.ceil(Math.sqrt(Math.pow(10, n))) - 1, 2));
}
 
// Driver code
var n = 4;
nDigitPerfectSquares(n);
 
// This code is contributed by rutvik_56.
</script>

Output:

1024 9801

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