# 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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