# Count squares with odd side length in Chessboard

Given a **N * N** chessboard, the task is to count the number of squares having the odd side length.**Example:**

Input:N = 3Output:10

9 squares are possible whose sides are 1

and a single square with side = 3

9 + 1 = 10Input:N = 8Output:120

**Approach:** For all odd numbers from **1** to **N** and then calculate the number of squares that can be formed having that odd side. For the **i ^{th}** side, the count of squares is equal to

**(N – i + 1)**. Further, add all such counts of squares.

^{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 odd length squares possible` `int` `count_square(` `int` `n)` `{` ` ` `// To store the required count` ` ` `int` `count = 0;` ` ` `// For all odd values of i` ` ` `for` `(` `int` `i = 1; i <= n; i = i + 2) {` ` ` `// Add the count of possible` ` ` `// squares of length i` ` ` `int` `k = n - i + 1;` ` ` `count += (k * k);` ` ` `}` ` ` `// Return the required count` ` ` `return` `count;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `N = 8;` ` ` `cout << count_square(N);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `class` `GFG {` ` ` `// Function to return the count` ` ` `// of odd length squares possible` ` ` `static` `int` `count_square(` `int` `n)` ` ` `{` ` ` `// To store the required count` ` ` `int` `count = ` `0` `;` ` ` `// For all odd values of i` ` ` `for` `(` `int` `i = ` `1` `; i <= n; i = i + ` `2` `) {` ` ` `// Add the count of possible` ` ` `// squares of length i` ` ` `int` `k = n - i + ` `1` `;` ` ` `count += (k * k);` ` ` `}` ` ` `// Return the required count` ` ` `return` `count;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `N = ` `8` `;` ` ` `System.out.println(count_square(N));` ` ` `}` `}` `// This code is contributed by Rajput-Ji` |

## Python3

`# Python implementation of the approach` `# Function to return the count` `# of odd length squares possible` `def` `count_square(n):` ` ` `# To store the required count` ` ` `count ` `=` `0` `;` ` ` `# For all odd values of i` ` ` `for` `i ` `in` `range` `(` `1` `, n ` `+` `1` `, ` `2` `):` ` ` `# Add the count of possible` ` ` `# squares of length i` ` ` `k ` `=` `n ` `-` `i ` `+` `1` `;` ` ` `count ` `+` `=` `(k ` `*` `k);` ` ` `# Return the required count` ` ` `return` `count;` `# Driver code` `N ` `=` `8` `;` `print` `(count_square(N));` `# This code has been contributed by 29AjayKumar` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG {` ` ` `// Function to return the count` ` ` `// of odd length squares possible` ` ` `static` `int` `count_square(` `int` `n)` ` ` `{` ` ` `// To store the required count` ` ` `int` `count = 0;` ` ` `// For all odd values of i` ` ` `for` `(` `int` `i = 1; i <= n; i = i + 2) {` ` ` `// Add the count of possible` ` ` `// squares of length i` ` ` `int` `k = n - i + 1;` ` ` `count += (k * k);` ` ` `}` ` ` `// Return the required count` ` ` `return` `count;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `N = 8;` ` ` `Console.WriteLine(count_square(N));` ` ` `}` `}` `// This code is contributed by Code_Mech.` |

## PHP

`<?php` `// PHP implementation of the approach` `// Function to return the count` `// of odd length squares possible` `function` `count_square(` `$n` `)` `{` ` ` `// To store the required count` ` ` `$count` `= 0;` ` ` `// For all odd values of i` ` ` `for` `(` `$i` `= 1; ` `$i` `<= ` `$n` `; ` `$i` `= ` `$i` `+ 2)` ` ` `{` ` ` `// Add the count of possible` ` ` `// squares of length i` ` ` `$k` `=` `$n` `- ` `$i` `+ 1;` ` ` `$count` `+= (` `$k` `* ` `$k` `);` ` ` `}` ` ` `// Return the required count` ` ` `return` `$count` `;` `}` `// Driver code` `$N` `= 8;` `echo` `count_square(` `$N` `);` `// This code is contributed by AnkitRai01` `?>` |

## Javascript

`<Script>` `// Javascript implementation of the approach` `// Function to return the count` `// of odd length squares possible` `function` `count_square(n)` `{` ` ` `// To store the required count` ` ` `let count = 0;` ` ` `// For all odd values of i` ` ` `for` `(let i = 1; i <= n; i = i + 2) {` ` ` `// Add the count of possible` ` ` `// squares of length i` ` ` `let k = n - i + 1;` ` ` `count += (k * k);` ` ` `}` ` ` `// Return the required count` ` ` `return` `count;` `}` `// Driver code` ` ` `let N = 8;` ` ` `document.write(count_square(N));` `</script>` |

**Output:**

120