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 

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 N^th term is (N² – (2 * N) + 2)
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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

Javascript

<script>

// Javascript implementation of the approach

// Function to return the count of
// squares that can be formed
function countSquares(n)
{
    return (Math.pow(n, 2) - (2 * n) + 2);
}

// Driver code
var n = 2;
document.write(countSquares(n));

</script>

Output:

2

Time Complexity: O(1)

Auxiliary Space: O(1)