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Find the number of squares inside the given square grid

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  • Last Updated : 08 Mar, 2022

Given a grid of side N * N, the task is to find the total number of squares that exist inside it. All squares selected can be of any length.
Examples: 
 

Input: N = 1 
Output:
 

Input: N = 2 
Output:
 

Input: N = 4 
Output: 30 
 

 

Approach 1: Taking a few examples, it can be observed that for a grid on size N * N, the number of squares inside it will be 12 + 22 + 32 + … + N2
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 number
// of squares inside an n*n grid
int cntSquares(int n)
{
    int squares = 0;
    for (int i = 1; i <= n; i++) {
        squares += pow(i, 2);
    }
    return squares;
}
 
// Driver code
int main()
{
    int n = 4;
 
    cout << cntSquares(4);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG {
 
    // Function to return the number
    // of squares inside an n*n grid
    static int cntSquares(int n)
    {
        int squares = 0;
        for (int i = 1; i <= n; i++) {
            squares += Math.pow(i, 2);
        }
        return squares;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 4;
 
        System.out.print(cntSquares(4));
    }
}

Python3




# Python3 implementation of the approach
 
# Function to return the number
# of squares inside an n*n grid
def cntSquares(n) :
 
    squares = 0;
    for i in range(1, n + 1) :
        squares += i ** 2;
 
    return squares;
 
# Driver code
if __name__ == "__main__" :
 
    n = 4;
 
    print(cntSquares(4));
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the number
    // of squares inside an n*n grid
    static int cntSquares(int n)
    {
        int squares = 0;
        for (int i = 1; i <= n; i++)
        {
            squares += (int)Math.Pow(i, 2);
        }
        return squares;
    }
 
    // Driver code
    public static void Main(String []args)
    {
        int n = 4;
 
        Console.Write(cntSquares(n));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
    // JavaScript implementation of the approach
     
    // Function to return the number
    // of squares inside an n*n grid
    function cntSquares(n)
    {
        let squares = 0;
        for (let i = 1; i <= n; i++)
        {
            squares += Math.pow(i, 2);
        }
        return squares;
    }
     
    let n = 4;
   
      document.write(cntSquares(n));
             
</script>

Output: 

30

 

Time Complexity: O(n)

Auxiliary Space: O(1)

Approach 2: By the use of direct formula. 
However, the sum \displaystyle \sum_{i=1}^{n}k^{2}     has the closed form (direct formula) \displaystyle \frac{n\left(n+1\right)\left(2n+1\right)}{6}     . Hence, we can employ this to calculate the sum in \displaystyle O(1)     time.
Below is the implementation of the above approach:
 

C++




// C++ implementation of the approach
#include <iostream>
 
using namespace std;
 
int cnt_squares (int n)
{
    /* Function to return the number
     of squares inside an n*n grid */
 
    return n * (n + 1) * (2 * n + 1) / 6;
}
 
// Driver code
int main()
{
    cout << cnt_squares (4) << endl;
 
    return 0;
}

Java




// Java implementation of the approach
class GFG {
    static int cntSquares (int n) {
        /* Function to return the number
        of squares inside an n*n grid */
     
        return n * (n + 1) * (2 * n + 1) / 6;
    }
 
    // Driver code
    public static void main(String args[]) {
        System.out.println (cntSquares(4));
    }
}

Python3




# Python3 implementation of the approach
 
"""
Function to return the number
of squares inside an n*n grid
"""
 
def cntSquares(n) :
    return int (n * (n + 1) * (2 * n + 1) / 6)
 
# Driver code
if __name__ == "__main__" :
    print (cntSquares (4));

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    /* Function to return the number
     of squares inside an n*n grid */
    static int cntSquares (int n)
    {
        return n * (n + 1) * (2 * n + 1) / 6;
    }
 
    // Driver code
    public static void Main (String[] args)
    {
        Console.Write (cntSquares (4));
    }
}

Javascript




<script>
    // Javascript implementation of the approach
     
    /* Function to return the number
     of squares inside an n*n grid */
    function cntSquares (n)
    {
        return n * (n + 1) * (2 * n + 1) / 6;
    }
     
    document.write(cntSquares(4));
 
// This code is contributed by divyeshrabadiya07.
</script>

Output: 

30

 

Time Complexity: O(1)

Auxiliary Space: O(1)


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