# Count of rooks that can attack each other out of K rooks placed on a N*N chessboard

• Last Updated : 07 Jan, 2022

Given pair of coordinates of K rooks on an N X N chessboard, the task is to count the number of rooks that can attack each other. Note: 1 <= K <= N*N
Examples:

Input: K = 2, arr[][] = { {2, 2}, {2, 3} }, N = 8
Output: 2
Explanation: Both the rooks can attack each other, because they are in the same row. Therefore, count of rooks that can attack each other is 2

Input: K = 1, arr[][] = { {4, 5} }, N = 4
Output: 0

Approach: The task can easily be solved using the fact that, 2 rooks can attack each other if they are either in the same row or in the same column, else they can’t attack each other.

Below is the implementation of the above code:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to count the number of attacking rooks``int` `willAttack(vector >& arr, ``int` `k, ``int` `N)``{``    ``int` `ans = 0;` `    ``for` `(``int` `i = 0; i < k; i++) {``        ``for` `(``int` `j = 0; j < k; j++) {` `            ``if` `(i != j) {` `                ``// Check if rooks are in same row``                ``// or same column``                ``if` `((arr[i] == arr[j])``                    ``|| (arr[i] == arr[j]))``                    ``ans++;``            ``}``        ``}``    ``}` `    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``vector > arr = { { 2, 2 }, { 2, 3 } };``    ``int` `K = 2, N = 8;``    ``cout << willAttack(arr, K, N);``    ``return` `0;``}`

## Java

 `import` `java.io.*;``import` `java.util.*;` `class` `Solution {``  ``static` `int` `willAttack(``int` `arr[][], ``int` `k, ``int` `N)``  ``{``    ``int` `ans = ``0``;` `    ``for` `(``int` `i = ``0``; i < k; i++) {``      ``for` `(``int` `j = ``0``; j < k; j++) {` `        ``if` `(i != j) {` `          ``// Check if rooks are in same row``          ``// or same column``          ``if` `((arr[i][``0``] == arr[j][``0``])``              ``|| (arr[i][``1``] == arr[j][``1``]))``            ``ans++;``        ``}``      ``}``    ``}` `    ``return` `ans;``  ``}``  ` `  ``// Driver code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``int``[][] arr = { { ``2``, ``2` `}, { ``2``, ``3` `} };``    ``int` `K = ``2``, N = ``8``;``    ``System.out.println(willAttack(arr, K, N));``  ``}``}` `// This code is contributed by dwivediyash.`

## Python3

 `# python program for the above approach` `# Function to count the number of attacking rooks``def` `willAttack(arr, k, N):` `    ``ans ``=` `0` `    ``for` `i ``in` `range``(``0``, k):``        ``for` `j ``in` `range``(``0``, k):` `            ``if` `(i !``=` `j):` `                ``# Check if rooks are in same row``                ``# or same column``                ``if` `((arr[i][``0``] ``=``=` `arr[j][``0``])``                        ``or` `(arr[i][``1``] ``=``=` `arr[j][``1``])):``                    ``ans ``+``=` `1` `    ``return` `ans` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[[``2``, ``2``], [``2``, ``3``]]``    ``K ``=` `2``    ``N ``=` `8``    ``print``(willAttack(arr, K, N))` `    ``# This code is contributed by rakeshsahni`

## C#

 `using` `System;``class` `Solution``{``  ``static` `int` `willAttack(``int``[,] arr, ``int` `k, ``int` `N)``  ``{``    ``int` `ans = 0;` `    ``for` `(``int` `i = 0; i < k; i++)``    ``{``      ``for` `(``int` `j = 0; j < k; j++)``      ``{` `        ``if` `(i != j)``        ``{` `          ``// Check if rooks are in same row``          ``// or same column``          ``if` `((arr[i, 0] == arr[j, 0])``              ``|| (arr[i, 1] == arr[j, 1]))``            ``ans++;``        ``}``      ``}``    ``}` `    ``return` `ans;``  ``}` `  ``// Driver code``  ``public` `static` `void` `Main()``  ``{``    ``int``[,] arr = { { 2, 2 }, { 2, 3 } };``    ``int` `K = 2, N = 8;``    ``Console.WriteLine(willAttack(arr, K, N));``  ``}``}` `// This code is contributed by gfgking`

## Javascript

 ``
Output
`2`

Time Complexity: O(K*K)
Auxiliary Space: O(1)

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