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 Nth term is (N2 – (2 * N) + 2)

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

2

Time Complexity: O(1)



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