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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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



Output:

1024 9801

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