# Count of squares that can be drawn without lifting the pencil

Given an integer N, the task is to find the count of squares of side 1 that can be drawn without lifting the pencil, starting at one corner of an N * N grid and never visiting an edge twice.

Input: N = 2
Output: 2

Input: N = 3
Output: 5

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

Approach: It can be observed that for the values of N = 1, 2, 3, …, a series will be formed as 1, 2, 5, 10, 17, 26, 37, 50, … whose Nth term is (N2 – (2 * N) + 2)

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of ` `// squares that can be formed ` `int` `countSquares(``int` `n) ` `{ ` `    ``return` `(``pow``(n, 2) - (2 * n) + 2); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 2; ` ` `  `    ``cout << countSquares(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` ` `  `// Function to return the count of ` `// squares that can be formed ` `static` `int` `countSquares(``int` `n) ` `{ ` `    ``return` `(``int``) (Math.pow(n, ``2``) - (``2` `* n) + ``2``); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String []args) ` `{ ` `    ``int` `n = ``2``; ` `    ``System.out.println(countSquares(n)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the count of  ` `# squares that can be formed  ` `def` `countSquares(n) : ` ` `  `    ``return` `(``pow``(n, ``2``) ``-` `(``2` `*` `n) ``+` `2``);  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``n ``=` `2``;  ` ` `  `    ``print``(countSquares(n));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `                     `  `class` `GFG ` `{ ` ` `  `    ``// Function to return the count of ` `    ``// squares that can be formed ` `    ``static` `int` `countSquares(``int` `n) ` `    ``{ ` `        ``return` `(``int``) (Math.Pow(n, 2) - (2 * n) + 2); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String []args) ` `    ``{ ` `        ``int` `n = 2; ` `        ``Console.WriteLine(countSquares(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```2
```

Time Complexity: O(1)

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