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 <bits/stdc++.h>using namespace std;// Function to count the number of attacking rooksint willAttack(vector<vector<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 Codeint main(){ vector<vector<int> > 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 rooksdef 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 Codeif __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
<script> // JavaScript code for the above approach // Function to count the number of attacking rooks function willAttack(arr, k, N) { let ans = 0; for (let i = 0; i < k; i++) { for (let 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 let arr = [[2, 2], [2, 3]]; let K = 2, N = 8; document.write(willAttack(arr, K, N));// This code is contributed by Potta Lokesh </script> |
Output
2
Time Complexity: O(K*K)
Auxiliary Space: O(1)
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