Check if players can meet on the same cell of the matrix in odd number of operations

Given four integers X, Y, A, and B. X and Y denotes the rows and the columns of a matrix such that (X, Y ≥ 2). Check if both the players can meet on the same cell of the matrix in an exactly odd number of operations by following the given conditions:

• Player 1 is at the top left corner of the matrix initially.
• Player 2 is at the bottom right of the matrix initially.
• In one operation, Player 1 can make exactly A steps, or Player 2 can make exactly B steps. In one step a player can move only left, right, up, or down cell from the current cell. 

Examples:

Input: X = 2, Y = 2, A = 2, B = 1 
Output: YES
Explanation: Operations are as follows:

• Operation 1: Player 1 can make exactly 2 steps and initially at the top left corner(1, 1). The sequence of two steps is as: (1, 1) -> (1, 2) -> (2, 2). Player 2 is initially present at the bottom right corner(N.M) = (2, 2). It can be verified that both players are present at (2, 2) now. The number of operations is 1, Which is exactly odd. Therefore, the output is YES.

Input: X = 2, Y = 2, A = 1, B = 2
Output: YES
Explanation: Operations are as follows:

• Operation 1: Player 1 moves to the right cell from the current position, Formally (1, 1) -> (1, 2).
• Operation 2: Player 2, First moves to the left cell from the current position(2, 2) and then right side to one step, (2, 2) -> (2, 1) -> (2, 2). Formally, returns to its initial position.    
• Operation 3: Player 1 is at currently (1, 2), Makes one step downwards from (1, 2) to (2, 2). It can be verified that now both the player are at (2, 2) in 3 operations, Which is odd. Therefore, the output is YES.

Input: X = 2, Y = 2, A = 1, B = 1
Output: NO
Explanation: It can be verified that it is not possible to meet on the same cell in an exactly odd number of operations. Hence the output is NO. 

Approach: Implement the idea below to solve the problem:

The problem is observation based and can be solved by using Mathematical concepts.

Steps were taken to solve the problem:

• Increment X by Y – 2 + A, Formally X+=Y-2+A.
• Increment A by B, Formally A += B.
• If (X % 2 == 1 && A % 2 == 0) then output YES else NO.

Below is the code to implement the above approach:

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// Method for output the answer
void Is_Possible_to_meet(int X, int Y, int A, int B)
{
     // Implementing idea for output the answer
     X += Y - 2 + A;
     A += B;
     cout<< (X % 2 == 1 && A % 2 == 0 ? "NO" : "YES") << endl;
}

int main() {

     // Input value of X and Y
     int X = 2, Y = 2;

     // Input value of A and B
     int A = 2, B = 1;

     // Function call
     Is_Possible_to_meet(X, Y, A, B);
}

Java

// Java code to implement the approach
import java.util.*;
class GFG {

    // Driver Function
    public static void main(String[] args)
    {

        // Input value of X and Y
        int X = 2, Y = 2;

        // Input value of A and B
        int A = 2, B = 1;

        // Function call
        Is_Possible_to_meet(X, Y, A, B);
    }

    // Method for output the answer
    static void Is_Possible_to_meet(int X, int Y, int A,
                                    int B)
    {

        // Implementing idea for output
        // the answer
        X += Y - 2 + A;
        A += B;
        System.out.println(
            X % 2 == 1 && A % 2 == 0 ? "NO" : "YES");
    }
}

C#

// C# code to implement the approach
using System;

public class GFG{

    // Method for output the answer
    static void Is_Possible_to_meet(int X, int Y, int A,
                                    int B)
    {

        // Implementing idea for output
        // the answer
        X += Y - 2 + A;
        A += B;
       Console.Write(
            X % 2 == 1 && A % 2 == 0 ? "NO" : "YES");
    }

    // Driver Function
    static public void Main (){

         // Input value of X and Y
        int X = 2, Y = 2;

        // Input value of A and B
        int A = 2, B = 1;

        // Function call
        Is_Possible_to_meet(X, Y, A, B);
    }
}

Javascript

// JS code to implement the approach

// Method for output the answer
function Is_Possible_to_meet( X, Y, A, B)
{
     // Implementing idea for output the answer
     X += Y - 2 + A;
     A += B;
     console.log(X % 2 == 1 && A % 2 == 0 ? "NO" : "YES");
}

// Input value of X and Y
let X = 2, Y = 2;

// Input value of A and B
let A = 2, B = 1;

// Function call
Is_Possible_to_meet(X, Y, A, B);

// this code is contributed by poojaagarwal2.

Python3

def Is_Possible_to_meet(X, Y, A, B):
    # Implementing idea for output the answer
    X += Y - 2 + A
    A += B
    print("NO" if X % 2 == 1 and A % 2 == 0 else "YES")

# Main function
if __name__ == '__main__':
    # Input value of X and Y
    X, Y = 2, 2

    # Input value of A and B
    A, B = 2, 1

    # Function call
    Is_Possible_to_meet(X, Y, A, B)

Output

YES

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