Given a 2D array knights[][] of size N * 2, with each row of the form { X, Y } representing the coordinates of knights, and an array pawn[] representing the coordinates of a pawn in an N * N board, the task is to find the count of knights present in the board that is attacking the pawn.
Examples:
Input: knights[][] = { { 0, 4 }, { 4, 5 }, { 1, 4 }, { 3, 1 } }, pawn[] = { 2, 3 }
Output: 2
Explanation:
The knights present at coordinate { { 0, 4 }, { 3, 1 } } are attacking the pawn.
Therefore, the required output is 2.Input: knights[][] = { { 4, 6 }, { 7, 5 }, { 5, 5 } }, pawn[] = { 6, 7 }
Output: 3
Explanation:
The knights present at coordinate { { 4, 6 }, { 7, 5 }, { 5, 5 } } are attacking the pawn.
Therefore, the required output is 3.
Approach: Follow the steps given below to solve the problem
- Initialize a variable, say cntKnights, to store the count of knights that are attacking the pawn.
- Traverse the knights[][] array using variable i and for every array element knights[i], check if the array { (knights[i][0] – pawn[0]), (knights[i][1] – pawn[1]) } is equal to either { 1, 2 } or { 2, 1 } or not. If found to be true, then increment the value of cntKnights by 1.
- Finally, print the value of cntKnights.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to count the knights that are// attacking the pawn in an M * M boardint cntKnightsAttackPawn(int knights[][2], int pawn[], int M){ // Stores count of knights that // are attacking the pawn int cntKnights = 0; // Traverse the knights[][] array for (int i = 0; i < M; i++) { // Stores absolute difference of X // co-ordinate of i-th knight and pawn int X = abs(knights[i][0] - pawn[0]); // Stores absolute difference of Y // co-ordinate of i-th knight and pawn int Y = abs(knights[i][1] - pawn[1]); // If X is 1 and Y is 2 or // X is 2 and Y is 1 if ((X == 1 && Y == 2) || (X == 2 && Y == 1)) { // Update cntKnights cntKnights++; } } return cntKnights;}// Driver Codeint main(){ int knights[][2] = { { 0, 4 }, { 4, 5 }, { 1, 4 }, { 3, 1 } }; int pawn[] = { 2, 3 }; // Stores total count of knights int M = sizeof(knights) / sizeof(knights[0]); cout << cntKnightsAttackPawn( knights, pawn, M); return 0;} |
Java
// Java program to implement // the above approach import java.io.*;import java.lang.Math;class GFG{// Function to count the knights that are // attacking the pawn in an M * M board static int cntKnightsAttackPawn(int knights[][], int pawn[], int M){ // Stores count of knights that // are attacking the pawn int cntKnights = 0; // Traverse the knights[][] array for(int i = 0; i < M; i++) { // Stores absolute difference of X // co-ordinate of i-th knight and pawn int X = Math.abs(knights[i][0] - pawn[0]); // Stores absolute difference of Y // co-ordinate of i-th knight and pawn int Y = Math.abs(knights[i][1] - pawn[1]); // If X is 1 and Y is 2 or // X is 2 and Y is 1 if ((X == 1 && Y == 2) || (X == 2 && Y == 1)) { // Update cntKnights cntKnights++; } } return cntKnights; }// Driver codepublic static void main(String[] args){ int[][] knights = { { 0, 4 }, { 4, 5 }, { 1, 4 }, { 3, 1 } }; int[] pawn = new int[]{2, 3}; // Stores total count of knights int M = knights.length; System.out.println(cntKnightsAttackPawn( knights, pawn, M));}}// This code is contributed by vandanakillari54935 |
Python3
# Python program to implement# the above approach# Function to count the knights that are# attacking the pawn in an M * M boarddef cntKnightsAttackPawn(knights, pawn, M): # Stores count of knights that # are attacking the pawn cntKnights = 0; # Traverse the knights array for i in range(M): # Stores absolute difference of X # co-ordinate of i-th knight and pawn X = abs(knights[i][0] - pawn[0]); # Stores absolute difference of Y # co-ordinate of i-th knight and pawn Y = abs(knights[i][1] - pawn[1]); # If X is 1 and Y is 2 or # X is 2 and Y is 1 if ((X == 1 and Y == 2) or (X == 2 and Y == 1)): # Update cntKnights cntKnights += 1; return cntKnights;# Driver codeif __name__ == '__main__': knights = [[0, 4], [4, 5], [1, 4], [3, 1]]; pawn = [2, 3]; # Stores total count of knights M = len(knights); print(cntKnightsAttackPawn(knights, pawn, M));# This code is contributed by Amit Katiyar |
C#
// C# program to implement // the above approach using System;class GFG { // Function to count the knights that are // attacking the pawn in an M * M board static int cntKnightsAttackPawn(int[,] knights, int[] pawn, int M) { // Stores count of knights that // are attacking the pawn int cntKnights = 0; // Traverse the knights[][] array for (int i = 0; i < M; i++) { // Stores absolute difference of X // co-ordinate of i-th knight and pawn int X = Math.Abs(knights[i, 0] - pawn[0]); // Stores absolute difference of Y // co-ordinate of i-th knight and pawn int Y = Math.Abs(knights[i, 1] - pawn[1]); // If X is 1 and Y is 2 or // X is 2 and Y is 1 if ((X == 1 && Y == 2) || (X == 2 && Y == 1)) { // Update cntKnights cntKnights++; } } return cntKnights; } // Driver code static void Main() { int[,] knights = {{ 0, 4 }, { 4, 5 }, { 1, 4 }, { 3, 1 }}; int[] pawn = {2, 3}; // Stores total count of knights int M = knights.GetLength(0); Console.WriteLine(cntKnightsAttackPawn(knights, pawn, M)); }}// This code is contributed by divyeshrabadiya07 |
Javascript
<script>// javascript program for the above approach// Function to count the knights that are // attacking the pawn in an M * M board function cntKnightsAttackPawn(knights, pawn, M){ // Stores count of knights that // are attacking the pawn let cntKnights = 0; // Traverse the knights[][] array for(let i = 0; i < M; i++) { // Stores absolute difference of X // co-ordinate of i-th knight and pawn let X = Math.abs(knights[i][0] - pawn[0]); // Stores absolute difference of Y // co-ordinate of i-th knight and pawn let Y = Math.abs(knights[i][1] - pawn[1]); // If X is 1 and Y is 2 or // X is 2 and Y is 1 if ((X == 1 && Y == 2) || (X == 2 && Y == 1)) { // Update cntKnights cntKnights++; } } return cntKnights; }// Driver Code let knights = [[ 0, 4 ], [ 4, 5 ], [ 1, 4 ], [ 3, 1 ]]; let pawn = [2, 3]; // Stores total count of knights let M = knights.length; document.write(cntKnightsAttackPawn( knights, pawn, M));</script> |
Output
2
Time Complexity: O(M), where M is the total count number of knights
Auxiliary Space: O(1)
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