Aquileo | Count Knights that can attack a given pawn in an N * N board

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 board
int 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 Code
int 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 code
public 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 board
def 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 code
if __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# 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

Time Complexity: O(M), where M is the total count number of knights
Auxiliary Space: O(1)

Count Knights that can attack a given pawn in an N * N board

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