Check if it is possible to reach destination in even number of steps in an Infinite Matrix

Given a source and destination in a matrix[][] of infinite rows and columns, the task is to find whether it is possible to reach the destination from the source in an even number of steps. Also, you can only move up, down, left, and right.

Examples:

Input: Source = {2, 1}, Destination = {1, 4}
Output: Yes

Input: Source = {2, 2}, Destination = {1, 4}
Output: No

Observation:

The observation is that if the steps required to reach the destination in the shortest path is even, then steps required in every other route to reach it will always be even. Also, there can be an infinite number of ways to reach the target point. Some paths to reach (4, 1) from (1, 2) in a 4 x 5 matrix are given below :

So our problem is reduced in finding the minimum number of steps required to reach the destination from source in a matrix, which can be calculated easily by simply taking the sum of the absolute values of the difference between the X coordinates and Y coordinates.

Below is the implementation of the above approach:

// C++ Program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check destination can be
// reached from source in even number of
// steps
void IsEvenPath(int Source[], int Destination[])
{
    // Coordinates differences
    int x_dif = abs(Source[0] - Destination[0]);
    int y_dif = abs(Source[1] - Destination[1]);

    // minimum number of steps required
    int minsteps = x_dif + y_dif;

    // Minsteps is even
    if (minsteps % 2 == 0)
        cout << "Yes";

    // Minsteps is odd
    else
        cout << "No";
}

// Driver Code
int main()
{
    // Given Input
    int Source[] = { 2, 1 };
    int Destination[] = { 1, 4 };

    // Function Call
    IsEvenPath(Source, Destination);

    return 0;
}

// Java program for the above approach
import java.lang.*;
import java.util.*;

class GFG{

// Function to check destination can be
// reached from source in even number of
// steps
static void IsEvenPath(int Source[], int Destination[])
{
    
    // Coordinates differences
    int x_dif = Math.abs(Source[0] - Destination[0]);
    int y_dif = Math.abs(Source[1] - Destination[1]);

    // Minimum number of steps required
    int minsteps = x_dif + y_dif;

    // Minsteps is even
    if (minsteps % 2 == 0)
        System.out.println("Yes");

    // Minsteps is odd
    else
        System.out.println("No");
}

// Driver code
public static void main(String[] args)
{
    
    // Given Input
    int Source[] = { 2, 1 };
    int Destination[] = { 1, 4 };

    // Function Call
    IsEvenPath(Source, Destination);
}
}

// This code is contributed by sanjoy_62

# Python3 program for the above approach

# Function to check destination can be
# reached from source in even number of
# steps
def IsEvenPath(Source, Destination):

    # Coordinates differences
    x_dif = abs(Source[0] - Destination[0])
    y_dif = abs(Source[1] - Destination[1])

    # Minimum number of steps required
    minsteps = x_dif + y_dif

    # Minsteps is even
    if (minsteps % 2 == 0):
        print("Yes")

    # Minsteps is odd
    else:
        print("No")

# Driver Code
if __name__ == '__main__':
    
    # Given Input
    Source = [ 2, 1 ]
    Destination = [ 1, 4 ]

    # Function Call
    IsEvenPath(Source, Destination)

# This code is contributed by mohit kumar 29

// C# program for the above approach
using System;

class GFG{

// Function to check destination can be
// reached from source in even number of
// steps
static void IsEvenPath(int[] Source, int[] Destination)
{
    
    // Coordinates differences
    int x_dif = Math.Abs(Source[0] - Destination[0]);
    int y_dif = Math.Abs(Source[1] - Destination[1]);

    // Minimum number of steps required
    int minsteps = x_dif + y_dif;

    // Minsteps is even
    if (minsteps % 2 == 0)
        Console.WriteLine("Yes");

    // Minsteps is odd
    else
        Console.WriteLine("No");
}

// Driver code
public static void Main(string[] args)
{
    // Given Input
    int[] Source = { 2, 1 };
    int[] Destination = { 1, 4 };

    // Function Call
    IsEvenPath(Source, Destination);
}
}

// This code is contributed by code_hunt.

 <script>
        // JavaScript Program for the above approach

        // Function to check destination can be
        // reached from source in even number of
        // steps
        function IsEvenPath(Source, Destination)
        {
        
            // Coordinates differences
            let x_dif = Math.abs(Source[0] - Destination[0]);
            let y_dif = Math.abs(Source[1] - Destination[1]);

            // minimum number of steps required
            let minsteps = x_dif + y_dif;

            // Minsteps is even
            if (minsteps % 2 == 0)
                document.write("Yes");

            // Minsteps is odd
            else
                document.write("No");
        }

        // Driver Code

        // Given Input
        let Source = [2, 1];
        let Destination = [1, 4];

        // Function Call
        IsEvenPath(Source, Destination);

    // This code is contributed by Potta Lokesh
    </script>

Yes

Time Complexity: O(1) // since no loop is used the algorithm takes up constant time to perform the operations
Auxiliary Space: O(1) // since no extra array is used so the space taken by the algorithm is constant