# Number of ways to place two queens on a N*N chess-board

• Last Updated : 20 Apr, 2021

Given an integer N denoting a N * N chess-board, the task is to count the number of ways to place two queens on the board such that, they do no attack each other.
Examples:

Input: N = 9
Output: 2184
Explanation:
There are 2184 ways to place two queens on 9 * 9 chess-board.
Input: N = 3
Output:
Explanation:
There are 8 ways to place two queens on 3 * 3 chess-board.

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In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students. Naive Approach: A simple solution will be to choose two every possible position for the two queens on the N * N matrix and check that they are not in horizontal, vertical, positive diagonal or negative diagonal. If yes then increment the count by 1.
Time Complexity: O(N4)
Efficient Approach: The idea is to use combinations to compute the possible positions of the queens such that they do not attack each other. A useful observation is that it is quite easy to calculate the number of positions that a single queen attacks. That is –

```Number of positions a queen attack =
(N - 1) + (N - 1) + (D - 1)

Here,
// First N-1 denotes positions in horizontal direction
// Second N-1 denotes positions in vertical direction
// D = Number of positions in
positive and negative diagonal```

If we do not place the queen on the last row and the last column then the answer will simply be the number of positions to place in a chessboard of (N-1)*(N-1), whereas if we place in the last column and last row then possible positions for queens will be 2*N – 1 and attacking at 3*(N – 1) positions. Therefore, the possible positions for the other queen will be N2 – 3*(N-1) – 1. Finally, there are (N-1)*(N-2) combinations where both queens are on the last row and last column. Therefore, the recurrence relation will be –

```Q(N) = Q(N-1) + (N2-3*(N-1)-1)-(N-1)*(N-2)

// By Induction
Q(N) = (N4)/2 - 5*(N3)/3 + 3*(N2)/2 - N/3```

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the` `// number of ways to place two ` `// queens on the N * N chess board`   `#include `   `#define ll long long` `using` `namespace` `std;`   `// Function to find number of valid ` `// positions for two queens in the` `// N * N chess board` `ll possiblePositions(ll n)` `{` `    ``ll term1 = ``pow``(n, 4);` `    ``ll term2 = ``pow``(n, 3);` `    ``ll term3 = ``pow``(n, 2);` `    ``ll term4 = n / 3;` `    ``ll ans = (``ceil``)(term1) / 2 - ` `             ``(``ceil``)(5 * term2) / 3 + ` `             ``(``ceil``)(3 * term3) / 2 - term4;` `    ``return` `ans;` `}`   `// Driver Code` `int` `main()` `{` `    ``ll n;` `    ``n = 3;` `    `  `    ``// Function Call` `    ``ll ans = possiblePositions(n);` `    ``cout << ans << endl;` `    ``return` `0;` `}`

## Java

 `// Java implementation to find the` `// number of ways to place two ` `// queens on the N * N chess board` `class` `GFG{`   `// Function to find number of valid ` `// positions for two queens in the` `// N * N chess board` `static` `double` `possiblePositions(``double` `n)` `{` `    ``double` `term1 = Math.pow(n, ``4``);` `    ``double` `term2 = Math.pow(n, ``3``);` `    ``double` `term3 = Math.pow(n, ``2``);` `    ``double` `term4 = n / ``3``;` `    ``double` `ans = (Math.ceil(term1 / ``2``)) - ` `                 ``(Math.ceil(``5` `* term2) / ``3``) + ` `                 ``(Math.ceil(``3` `* term3) / ``2``) - term4;`   `    ``return` `(``long``)ans;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``double` `n;` `    ``n = ``3``;` `    `  `    ``// Function Call` `    ``double` `ans = possiblePositions(n);` `    ``System.out.print(ans + ``"\n"``);` `}` `}`   `// This code is contributed by sapnasingh4991`

## Python3

 `# Python3 implementation to find the ` `# number of ways to place two ` `# queens on the N * N chess board ` `import` `math `   `# Function to find number of valid ` `# positions for two queens in the ` `# N * N chess board ` `def` `possiblePositions(n): ` `    `  `    ``term1 ``=` `pow``(n, ``4``); ` `    ``term2 ``=` `pow``(n, ``3``); ` `    ``term3 ``=` `pow``(n, ``2``); ` `    ``term4 ``=` `n ``/` `3``; ` `    `  `    ``ans ``=` `((math.ceil(term1)) ``/` `2` `-` `           ``(math.ceil(``5` `*` `term2)) ``/` `3` `+` `           ``(math.ceil(``3` `*` `term3)) ``/` `2` `-` `term4); ` `           `  `    ``return` `ans; `   `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    `  `    ``n ``=` `3`   `    ``# Function call` `    ``ans ``=` `possiblePositions(n) ` `    `  `    ``print``(``int``(ans)) `   `# This code is contributed by jana_sayantan`

## C#

 `// C# implementation to find the` `// number of ways to place two ` `// queens on the N * N chess board` `using` `System;`   `class` `GFG{`   `// Function to find number of valid ` `// positions for two queens in the` `// N * N chess board` `static` `double` `possiblePositions(``double` `n)` `{` `    ``double` `term1 = Math.Pow(n, 4);` `    ``double` `term2 = Math.Pow(n, 3);` `    ``double` `term3 = Math.Pow(n, 2);` `    ``double` `term4 = n / 3;` `    ``double` `ans = (Math.Ceiling(term1 / 2)) - ` `                 ``(Math.Ceiling(5 * term2) / 3) + ` `                 ``(Math.Ceiling(3 * term3) / 2) - term4;`   `    ``return` `(``long``)ans;` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``double` `n;` `    ``n = 3;` `    `  `    ``// Function Call` `    ``double` `ans = possiblePositions(n);` `    ``Console.Write(ans + ``"\n"``);` `}` `}`   `// This code is contributed by Amit Katiyar`

## Javascript

 ``

Output:

`8`

Time Complexity: O(1)

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