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Count of squares that can be drawn without lifting the pencil

  • Last Updated : 06 Apr, 2021

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:
 

Input: N = 3 
Output:
 

 



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 <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)
 




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