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

• Last Updated : 24 Feb, 2022

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:
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:
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 ``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

 ``

Output:

`5`

Time Complexity: O(N * M)

Auxiliary Space: O(1)

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