# Sum of the count of number of adjacent squares in an M X N grid

Given an **M × N** matrix. The task is to count the number of adjacent cells and calculate their sum.

Two cells are said to be connected if they are adjacent to each other horizontally, vertically, or diagonally.**Examples :**

Input :m = 2, n = 2Output :12Input :m = 3, n = 2Output:22

See the below diagram where numbers written on it denotes number of adjacent squares.

**Approach:**

In a m X n grid there can be 3 cases:

- Corner cells touch 3 cells, and there are always 4 corner cells.
- Edge cells touch 5 cells, and there are always 2 * (m+n-4) edge cells.
- Interior cells touch 8 cells, and there are always (m-2) * (n-2) interior cells.

Therefore,

Sum = 3*4 + 5*2*(m+n-4) + 8*(m-2)*(n-2) = 8mn - 6m - 6n +4

Below is the implementation of the above approach:

## C++

`// C++ implementation of the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// function to calculate the sum of all cells adjacent value` `int` `sum(` `int` `m, ` `int` `n)` `{` ` ` `return` `8 * m * n - 6 * m - 6 * n + 4;` `}` `// Driver program to test above` `int` `main()` `{` ` ` `int` `m = 3, n = 2;` ` ` `cout << sum(m, n);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the above approach` `class` `GFG` `{` ` ` ` ` `// function to calculate the sum` ` ` `// of all cells adjacent value` ` ` `static` `int` `sum(` `int` `m, ` `int` `n)` ` ` `{` ` ` `return` `8` `* m * n - ` `6` `* m - ` `6` `* n + ` `4` `;` ` ` `}` ` ` ` ` `// Driver Code` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `int` `m = ` `3` `, n = ` `2` `;` ` ` `System.out.println(sum(m, n));` ` ` `}` `}` `// This Code is contributed by AnkitRai01` |

## Python3

`# Python3 implementation of the above approach` `# function to calculate the sum` `# of all cells adjacent value` `def` `summ(m, n):` ` ` `return` `8` `*` `m ` `*` `n ` `-` `6` `*` `m ` `-` `6` `*` `n ` `+` `4` `# Driver Code` `m ` `=` `3` `n ` `=` `2` `print` `(summ(m, n))` `# This code is contributed by Mohit Kumar` |

## C#

`// C# implementation of the above approach` `using` `System;` `class` `GFG` `{` ` ` ` ` `// function to calculate the sum` ` ` `// of all cells adjacent value` ` ` `static` `int` `sum(` `int` `m, ` `int` `n)` ` ` `{` ` ` `return` `8 * m * n - 6 * m - 6 * n + 4;` ` ` `}` ` ` ` ` `// Driver Code` ` ` `public` `static` `void` `Main (String []args)` ` ` `{` ` ` `int` `m = 3, n = 2;` ` ` `Console.WriteLine(sum(m, n));` ` ` `}` `}` `// This code is contributed by andrew1234` |

## Javascript

`<script>` `// Javascript implementation of the above approach` `// function to calculate the sum of all cells adjacent value` `function` `sum(m, n)` `{` ` ` `return` `8 * m * n - 6 * m - 6 * n + 4;` `}` `// Driver program to test above` `var` `m = 3, n = 2;` `document.write(sum(m, n));` `</script>` |

**Output:**

22

**Time Complexity:**

**Auxiliary Space: **O(1)