Number of cells in matrix which are equidistant from given two points

Given a matrix of N rows and M columns, given two points on the matrix; the task is to count the number of cells that are equidistant from given two points. Any traversal either in the horizontal direction or vertical direction or both ways is considered valid but the diagonal path is not valid.
Examples:

Input: 5 5 
2 4 
5 3 
Output: 5 
Explanation:
Out of all cells, these are the points (3, 1);(3, 2);(3, 3);(4, 4);(4, 5) 
which satisfy given condition.
Input: 4 3 
2 3 
4 1 
Output: 4 
Explanation:
Out of all cells, these are the points (1, 1);(2, 1);(3, 2);(4, 3) 
which satisfy given condition. 

Approach: 

1. Every cell of the matrix is traversed.
2. Let ‘A’ be the distance between current cell and first point and similarly ‘B’ be the distance between current cell and second point.
3. Distance between two points is calculated using Manhattan distance. If A & B are equal, the count is incremented.

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
int numberOfPoints(int N, int M, int x1,
                   int y1, int x2, int y2)
{

    // Initializing count
    int count = 0;

    // Traversing through rows.
    for (int i = 1; i <= N; i++) {

        // Traversing through columns.
        for (int j = 1; j <= M; j++) {

            // By using Manhattan Distance, the distance between
            // the current point to given two points is calculated.

            // If distances are equal
            // the count is incremented by 1.
            if (abs(i - x1) + abs(j - y1)
                == abs(i - x2) + abs(j - y2))
                count++;
        }
    }

    return count;
}

// Driver Code
int main()
{
    int n = 5;
    int m = 5;
    int x1 = 2;
    int y1 = 4;
    int x2 = 5;
    int y2 = 3;

    cout << numberOfPoints(n, m, x1, y1, x2, y2);
}

Java

//Java implementation of the above approach
import java.util.*;
import java.lang.Math;
public class GFG {
    public static int numberOfPoints(int N, int M, int x1,
                                     int y1, int x2, int y2)
    {
        int count = 0, i, j;

        // Traversing through rows.
        for (i = 1; i <= N; i++) {

            // Traversing through columns.
            for (j = 1; j <= M; j++) {

                // By using Manhattan Distance, distance between
                // current point to given two points is calculated

                // If distances are equal
                // the count is incremented by 1.
                if (Math.abs(i - x1) + Math.abs(j - y1)
                    == Math.abs(i - x2) + Math.abs(j - y2))
                    count += 1;
            }
        }
        return count;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int n = 5;
        int m = 5;
        int x1 = 2;
        int y1 = 4;
        int x2 = 5;
        int y2 = 3;

        System.out.println(numberOfPoints(n, m, x1, y1, x2, y2));
    }
}

Python

# Python implementation of the above approach
def numberPoints(N, M, x1, y1, x2, y2):

    # Initializing count = 0
    count = 0

    # Traversing through rows.
    for i in range(1, N + 1):

        # Traversing through columns.
        for j in range(1, M + 1):

            # By using Manhattan Distance,
            # distance between current point to
            # given two points is calculated

            # If distances are equal the
            # count is incremented by 1.
            if (abs(i - x1)+abs(j - y1)) == (abs(i - x2)+abs(j - y2)):
                count += 1

    return count

# Driver Code
N = 5
M = 5
x1 = 2
y1 = 4
x2 = 5
y2 = 3
print(numberPoints(N, M, x1, y1, x2, y2))

C#

// C# implementation of the above approach
using System;

class GFG
{

    static int numberOfPoints(int N, int M, int x1,
                                    int y1, int x2, int y2)
    {
        int count = 0, i, j;

        // Traversing through rows.
        for (i = 1; i <= N; i++)
        {

            // Traversing through columns.
            for (j = 1; j <= M; j++)
            {

                // By using Manhattan Distance, distance between
                // current point to given two points is calculated

                // If distances are equal
                // the count is incremented by 1.
                if (Math.Abs(i - x1) + Math.Abs(j - y1)
                    == Math.Abs(i - x2) + Math.Abs(j - y2))

                    count += 1;
            }
        }
        return count;
    }

    // Driver Code
    public static void Main()
    {
        int n = 5;
        int m = 5;
        int x1 = 2;
        int y1 = 4;
        int x2 = 5;
        int y2 = 3;

        Console.WriteLine(numberOfPoints(n, m, x1, y1, x2, y2));
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>
// javascript implementation of the above approach

    function numberOfPoints(N , M , x1 , y1 , x2 , y2)
    {
        var count = 0, i, j;

        // Traversing through rows.
        for (i = 1; i <= N; i++) {

            // Traversing through columns.
            for (j = 1; j <= M; j++) {

                // By using Manhattan Distance, distance between
                // current point to given two points is calculated

                // If distances are equal
                // the count is incremented by 1.
                if (Math.abs(i - x1) + Math.abs(j - y1) == Math.abs(i - x2) + Math.abs(j - y2))
                    count += 1;
            }
        }
        return count;
    }

    // Driver Code
        var n = 5;
        var m = 5;
        var x1 = 2;
        var y1 = 4;
        var x2 = 5;
        var y2 = 3;

        document.write(numberOfPoints(n, m, x1, y1, x2, y2));

// This code is contributed by Rajput-Ji
</script>

Output:

5

Time Complexity: O(N * M)

Auxiliary Space: O(1)