Puzzle | Program to find number of squares in a chessboard

Puzzle: You are provided with a chessboard and are asked to find the number of squares in it. A chessboard is a board with 8 x 8 grids in it as represented below. 

Solution: Looking closely at the chessboard we can see that in addition to the 1 x 1 square, there can be a combination of 2 x 2, 3 x 3, 4 x 4, 5 x 5, 6 x 6, 7 x 7, and 8 x 8 squares too. To get the total number of squares we need to find all the squares formed. 

1 x 1: 8 * 8 = 64 squares.
2 x 2: 7 * 7 = 49 squares.
3 x 3: 6 * 6 = 36 squares.
4 x 4: 5 * 5 = 25 squares.
5 x 5: 4 * 4 = 16 squares.
6 x 6: 3 * 3 = 9 squares.
7 x 7: 2 * 2 = 4 squares.
8 x 8: 1 * 1 = 1 square.

Therefore, we have in all = 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 204 squares in a chessboard.

General Process
Given an n x n grid, count squares in it. 

Examples:

Input: n = 2
Output: 5 (4 squares of 1 unit + 1 square of 2 units)

Input: n = 3
Output: 14 (9 squares of 1 unit + 4 square of 2 units 
                                + 1 square of 1 unit) 

For a grid of size n*n the total number of squares formed are: 

1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1) / 6 

Below is the implementation of the above formula. Since the value of n*(n+1)*(2n+1) can cause overflow for large values of n, below are some interesting tricks used in the program. 

1. long int is used in return.
2. n * (n + 1) / 2 is evaluated first as the value n*(n+1) will always be a multiple of 2.

Note that overflow may still happen, but the above tricks just reduce the chances of an overflow.  

C++

// C++  find number of squares in a
// chessboard
#include <bits/stdc++.h>
using namespace std;

// Function to return count of squares;
long long int countSquares(int n)
{
  return (n * (n + 1) / 2) * (2 * n + 1) / 3;
}

// Driver Code
int main()
{

  int n = 4;
  cout << countSquares(n);
  return 0;
}

//vermay87    `

Java

// Java  find number of squares in a
// chessboard


class GFG
{
    // Function to return count of squares;
    static int countSquares(int n)
    {
        // A better way to write n*(n+1)*(2n+1)/6
        return (n * (n + 1) / 2) * (2 * n + 1) / 3;
    }

    // Driver code
    public static void main (String[] args)
    {
        int n = 3;
        System.out.println("Count of squares is "
                           +countSquares(n));
    }
}

// This code is contributed by Anant Agarwal.

Python3

# python code to find number
# of squares in a chessboard

# Function to return count
# of squares;
def countSquares(n):

    # better way to write
    # n*(n+1)*(2n+1)/6
    return ((n * (n + 1) / 2)
           * (2 * n + 1) / 3)

# Driver code
n = 4
print("Count of squares is ",
              countSquares(n))

# This code is contributed by sam007.

C#

// C# find number of squares in a
// chessboard
using System;

public class GFG {

    static int countSquares(int n)
    {
        // A better way to write
        // n*(n+1)*(2n+1)/6
        return (n * (n + 1) / 2)
                * (2 * n + 1) / 3;
    }

    // Driver code
    public static void Main ()
    {
        int n = 4;
        Console.WriteLine("Count of"
                     + "squares is "
                 + countSquares(n));
    }
}

// This code is contributed by Sam007.

PHP

<?php
// PHP program to find number
// of squares in a chessboard

// Function to return
// count of squares;
function countSquares($n)
{
    // A better way to
    // write n*(n+1)*(2n+1)/6
    return ($n * ($n + 1) / 2) *
               (2 * $n + 1) / 3;
}

// Driver Code
$n = 4;
echo "Count of squares is " ,
            countSquares($n);

// This code is contributed
// by nitin mittal.
?>

Javascript

<script>

// Java  find number of squares in a
// chessboard

// Function to return count of squares;
function countSquares( n)
{
    // A better way to write n*(n+1)*(2n+1)/6
    return (n * (n + 1) / 2) * (2*n + 1) / 3;
}

// Driver Code

let n = 4;
document.write("Count of squares is " + countSquares(n));

</script>

Output:

Count of squares is 30

Time Complexity: O(1)

Auxiliary Space: O(1)

This article is contributed by Rishabh. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above