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

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

Time Complexity: $O(1)$

Auxiliary Space: O(1)

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