# Maximum product of 4 adjacent elements in matrix

• Difficulty Level : Easy
• Last Updated : 30 Jun, 2021

Given a square matrix, find the maximum product of four adjacent elements of the matrix. The adjacent elements of the matrix can be top, down, left, right, diagonal, or anti-diagonal. The four or more numbers should be adjacent to each other.

Note: n should be greater than or equal to 4 i.e n >= 4

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Examples :

```Input : n = 4
{{6, 2, 3 4},
{5, 4, 3, 1},
{7, 4, 5, 6},
{8, 3, 1, 0}}

Output : 1680

Explanation:
Multiplication of 6 5 7 8 produces maximum
result and all element are adjacent to
each other in one direction

Input : n = 5
{{1, 2, 3, 4, 5},
{6, 7, 8, 9, 1},
{2, 3, 4, 5, 6},
{7, 8, 9, 1, 0},
{9, 6, 4, 2, 3}}

Output: 3024

Explanation:
Multiplication of 6 7 8 9 produces maximum
result and all elements are adjacent to
each other in one direction.```

Approach:

1. Group 4 elements that are adjacent to each other in each row and calculate their maximum result.
2. Group 4 elements that are adjacent to each other in each column and calculate their maximum results.
3. Group 4 elements that are adjacent to each other in diagonal and calculate their maximum results.
4. Group 4 elements that are adjacent to each other in anti-diagonal and calculate their maximum results.
5. Compare all calculated maximum results.

Below is the implementation of the above approach:

## C++

 `// C++ program to find out the maximum product``// in the matrix which four elements are``// adjacent to each other in one direction``#include ``using` `namespace` `std;` `const` `int` `n = 5;` `// function to find max product``int` `FindMaxProduct(``int` `arr[][n], ``int` `n)``{``    ``int` `max = 0, result;` `    ``// iterate the rows.``    ``for` `(``int` `i = 0; i < n; i++)``    ``{` `        ``// iterate the columns.``        ``for` `(``int` `j = 0; j < n; j++)``        ``{` `            ``// check the maximum product``            ``// in horizontal row.``            ``if` `((j - 3) >= 0)``            ``{``                ``result = arr[i][j] * arr[i][j - 1] *``                    ``arr[i][j - 2] * arr[i][j - 3];``                ` `                ``if` `(max < result)``                    ``max = result;``            ``}` `            ``// check the maximum product``            ``// in vertical row.``            ``if` `((i - 3) >= 0)``            ``{``                ``result = arr[i][j] * arr[i - 1][j] *``                    ``arr[i - 2][j] * arr[i - 3][j];``                ` `                ``if` `(max < result)``                    ``max = result;``            ``}` `            ``// check the maximum product in``            ``// diagonal (going through down - right)``            ``if` `((i - 3) >= 0 && (j - 3) >= 0)``            ``{``                ``result = arr[i][j] * arr[i - 1][j - 1] *``                    ``arr[i - 2][j - 2] * arr[i - 3][j - 3];``                ` `                ``if` `(max < result)``                    ``max = result;``            ``}``            ` `            ``// check the maximum product in``            ``// diagonal (going through up - right)``            ``if` `((i - 3) >= 0 && (j - 1) <= 0)``            ``{``                ``result = arr[i][j] * arr[i - 1][j + 1] *``                    ``arr[i - 2][j + 2] * arr[i - 3][j + 3];``    ` `                ``if` `(max < result)``                    ``max = result;``            ``}``        ``}``    ``}` `    ``return` `max;``}` `// Driver code``int` `main()``{` `    ``/* int arr[] = {{6, 2, 3, 4},``                    ``{5, 4, 3, 1},``                    ``{7, 4, 5, 6},``                    ``{8, 3, 1, 0}};*/``    ``/* int arr[] = {{1, 2, 1, 3, 4},``                    ``{5, 6, 3, 9, 2},``                    ``{7, 8, 8, 1, 2},``                    ``{1, 0, 7, 9, 3},``                    ``{3, 0, 8, 4, 9}};*/``                        ` `    ``int` `arr[] = {{1, 2, 3, 4, 5},``                    ``{6, 7, 8, 9, 1},``                    ``{2, 3, 4, 5, 6},``                    ``{7, 8, 9, 1, 0},``                    ``{9, 6, 4, 2, 3}};` `    ``cout << FindMaxProduct(arr, n);``    ``return` `0;``}`

## Java

 `// Java program to find out the``// maximum product in the matrix``// which four elements are adjacent``// to each other in one direction``class` `GFG {``    ``static` `final` `int` `n = ``5``;` `    ``// function to find max product``    ``static` `int` `FindMaxProduct(``int` `arr[][], ``int` `n)``    ``{``        ``int` `max = ``0``, result;` `        ``// iterate the rows.``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``// iterate the columns.``            ``for` `(``int` `j = ``0``; j < n; j++)``            ``{``                ``// check the maximum product``                ``// in horizontal row.``                ``if` `((j - ``3``) >= ``0``)``                ``{``                    ``result = arr[i][j] * arr[i][j - ``1``]``                             ``* arr[i][j - ``2``]``                             ``* arr[i][j - ``3``];``                    ``if` `(max < result)``                        ``max = result;``                ``}` `                ``// check the maximum product``                ``// in vertical row.``                ``if` `((i - ``3``) >= ``0``)``                ``{``                    ``result = arr[i][j] * arr[i - ``1``][j]``                             ``* arr[i - ``2``][j]``                             ``* arr[i - ``3``][j];` `                    ``if` `(max < result)``                        ``max = result;``                ``}` `                ``// check the maximum product in``                ``// diagonal (going through down - right)``                ``if` `((i - ``3``) >= ``0` `&& (j - ``3``) >= ``0``)``                ``{``                    ``result = arr[i][j] * arr[i - ``1``][j - ``1``]``                             ``* arr[i - ``2``][j - ``2``]``                             ``* arr[i - ``3``][j - ``3``];` `                    ``if` `(max < result)``                        ``max = result;``                ``}` `                ``// check the maximum product in``                ``// diagonal (going through up - right)``                ``if` `((i - ``3``) >= ``0` `&& (j - ``1``) <= ``0``)``                ``{``                    ``result = arr[i][j] * arr[i - ``1``][j + ``1``]``                             ``* arr[i - ``2``][j + ``2``]``                             ``* arr[i - ``3``][j + ``3``];` `                    ``if` `(max < result)``                        ``max = result;``                ``}``            ``}``        ``}` `        ``return` `max;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{` `        ``/* int arr[] = {{6, 2, 3, 4},``                           ``{5, 4, 3, 1},``                           ``{7, 4, 5, 6},``                           ``{8, 3, 1, 0}};*/``        ``/* int arr[] = {{1, 2, 1, 3, 4},``                           ``{5, 6, 3, 9, 2},``                           ``{7, 8, 8, 1, 2},``                           ``{1, 0, 7, 9, 3},``                           ``{3, 0, 8, 4, 9}};*/` `        ``int` `arr[][] = { { ``1``, ``2``, ``3``, ``4``, ``5` `},``                        ``{ ``6``, ``7``, ``8``, ``9``, ``1` `},``                        ``{ ``2``, ``3``, ``4``, ``5``, ``6` `},``                        ``{ ``7``, ``8``, ``9``, ``1``, ``0` `},``                        ``{ ``9``, ``6``, ``4``, ``2``, ``3` `} };` `        ``System.out.print(FindMaxProduct(arr, n));``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python3 program to find out the maximum``# product in the matrix which four elements``# are adjacent to each other in one direction``n ``=` `5` `# function to find max product``def` `FindMaxProduct(arr, n):` `    ``max` `=` `0` `    ``# iterate the rows.``    ``for` `i ``in` `range``(n):` `        ``# iterate the columns.``        ``for` `j ``in` `range``( n):` `            ``# check the maximum product``            ``# in horizontal row.``            ``if` `((j ``-` `3``) >``=` `0``):``                ``result ``=` `(arr[i][j] ``*` `arr[i][j ``-` `1``] ``*``                          ``arr[i][j ``-` `2``] ``*` `arr[i][j ``-` `3``])``                ` `                ``if` `(``max` `< result):``                    ``max` `=` `result` `            ``# check the maximum product``            ``# in vertical row.``            ``if` `((i ``-` `3``) >``=` `0``) :``                ``result ``=` `(arr[i][j] ``*` `arr[i ``-` `1``][j] ``*``                          ``arr[i ``-` `2``][j] ``*` `arr[i ``-` `3``][j])``                ` `                ``if` `(``max` `< result):``                    ``max` `=` `result` `            ``# check the maximum product in``            ``# diagonal going through down - right``            ``if` `((i ``-` `3``) >``=` `0` `and` `(j ``-` `3``) >``=` `0``):``                ``result ``=` `(arr[i][j] ``*` `arr[i ``-` `1``][j ``-` `1``] ``*``                          ``arr[i ``-` `2``][j ``-` `2``] ``*` `arr[i ``-` `3``][j ``-` `3``])``                ` `                ``if` `(``max` `< result):``                    ``max` `=` `result` `            ``# check the maximum product in``            ``# diagonal going through up - right``            ``if` `((i ``-` `3``) >``=` `0` `and` `(j ``-` `1``) <``=` `0``):``                ``result ``=` `(arr[i][j] ``*` `arr[i ``-` `1``][j ``+` `1``] ``*``                          ``arr[i ``-` `2``][j ``+` `2``] ``*` `arr[i ``-` `3``][j ``+` `3``])` `                ``if` `(``max` `< result):``                    ``max` `=` `result` `    ``return` `max` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    `  `    ``# int arr[] = {{6, 2, 3, 4},``    ``#                  {5, 4, 3, 1},``    ``#                  {7, 4, 5, 6},``    ``#                  {8, 3, 1, 0}};``    ``# int arr[] = {{1, 2, 1, 3, 4},``    ``#                  {5, 6, 3, 9, 2},``    ``#                  {7, 8, 8, 1, 2},``    ``#                  {1, 0, 7, 9, 3},``    ``#                  {3, 0, 8, 4, 9}};``                        ` `    ``arr ``=` `[[``1``, ``2``, ``3``, ``4``, ``5``],``           ``[``6``, ``7``, ``8``, ``9``, ``1``],``           ``[``2``, ``3``, ``4``, ``5``, ``6``],``           ``[``7``, ``8``, ``9``, ``1``, ``0``],``            ``[``9``, ``6``, ``4``, ``2``, ``3``]]` `    ``print``(FindMaxProduct(arr, n))` `# This code is contributed by ita_c`

## C#

 `// C# program to find out the``// maximum product in the matrix``// which four elements are adjacent``// to each other in one direction``using` `System;` `public` `class` `GFG {` `    ``static` `int` `n = 5;` `    ``// Function to find max product``    ``static` `int` `FindMaxProduct(``int``[, ] arr, ``int` `n)``    ``{``        ``int` `max = 0, result;` `        ``// iterate the rows``        ``for` `(``int` `i = 0; i < n; i++) {` `            ``// iterate the columns``            ``for` `(``int` `j = 0; j < n; j++) {` `                ``// check the maximum product``                ``// in horizontal row.``                ``if` `((j - 3) >= 0) {` `                    ``result = arr[i, j] * arr[i, j - 1]``                             ``* arr[i, j - 2]``                             ``* arr[i, j - 3];` `                    ``if` `(max < result)``                        ``max = result;``                ``}` `                ``// check the maximum product``                ``// in vertical row.``                ``if` `((i - 3) >= 0) {``                    ``result = arr[i, j] * arr[i - 1, j]``                             ``* arr[i - 2, j]``                             ``* arr[i - 3, j];` `                    ``if` `(max < result)``                        ``max = result;``                ``}` `                ``// check the maximum product in``                ``// diagonal going through down - right``                ``if` `((i - 3) >= 0 && (j - 3) >= 0) {``                    ``result = arr[i, j] * arr[i - 1, j - 1]``                             ``* arr[i - 2, j - 2]``                             ``* arr[i - 3, j - 3];` `                    ``if` `(max < result)``                        ``max = result;``                ``}` `                ``// check the maximum product in``                ``// diagonal going through up - right``                ``if` `((i - 3) >= 0 && (j - 1) <= 0) {``                    ``result = arr[i, j] * arr[i - 1, j + 1]``                             ``* arr[i - 2, j + 2]``                             ``* arr[i - 3, j + 3];` `                    ``if` `(max < result)``                        ``max = result;``                ``}``            ``}``        ``}` `        ``return` `max;``    ``}` `    ``// Driver Code``    ``static` `public` `void` `Main()``    ``{``        ``int``[, ] arr = { { 1, 2, 3, 4, 5 },``                        ``{ 6, 7, 8, 9, 1 },``                        ``{ 2, 3, 4, 5, 6 },``                        ``{ 7, 8, 9, 1, 0 },``                        ``{ 9, 6, 4, 2, 3 } };` `        ``Console.Write(FindMaxProduct(arr, n));``    ``}``}` `// This code is contributed by Shrikant13`

## PHP

 `= 0)``            ``{``                ``\$result` `= ``\$arr``[``\$i``][``\$j``] *``                          ``\$arr``[``\$i``][``\$j` `- 1] *``                          ``\$arr``[``\$i``][``\$j` `- 2] *``                          ``\$arr``[``\$i``][``\$j` `- 3];``                ` `                ``if` `(``\$max` `< ``\$result``)``                    ``\$max` `= ``\$result``;``            ``}` `            ``// check the maximum product``            ``// in vertical row.``            ``if` `((``\$i` `- 3) >= 0)``            ``{``                ``\$result` `= ``\$arr``[``\$i``][``\$j``] *``                          ``\$arr``[``\$i` `- 1][``\$j``] *``                          ``\$arr``[``\$i` `- 2][``\$j``] *``                          ``\$arr``[``\$i` `- 3][``\$j``];``                ` `                ``if` `(``\$max` `< ``\$result``)``                    ``\$max` `= ``\$result``;``            ``}` `            ``// check the maximum product in``            ``// diagonal going through down - right``            ``if` `((``\$i` `- 3) >= 0 ``and` `(``\$j` `- 3) >= 0)``            ``{``                ``\$result` `= ``\$arr``[``\$i``][``\$j``] *``                          ``\$arr``[``\$i` `- 1][``\$j` `- 1] *``                          ``\$arr``[``\$i` `- 2][``\$j` `- 2] *``                          ``\$arr``[``\$i` `- 3][``\$j` `- 3];``                ` `                ``if` `(``\$max` `< ``\$result``)``                    ``\$max` `= ``\$result``;``            ``}``            ` `            ``// check the maximum product in``            ``// diagonal going through up - right``            ``if` `((``\$i` `- 3) >= 0 ``and` `(``\$j` `- 1) <= 0)``            ``{``                ``\$result` `= ``\$arr``[``\$i``][``\$j``] *``                          ``\$arr``[``\$i` `- 1][``\$j` `+ 1] *``                          ``\$arr``[``\$i` `- 2][``\$j` `+ 2] *``                          ``\$arr``[``\$i` `- 3][``\$j` `+ 3];``                ` `                ``if` `(``\$max` `< ``\$result``)``                    ``\$max` `= ``\$result``;``            ``}``            ` `        ``}``    ``}` `    ``return` `\$max``;``}``    ` `    ``// Driver Code                       ``    ``\$arr` `= ``array``(``array``(1, 2, 3, 4, 5),``                 ``array``(6, 7, 8, 9, 1),``                 ``array``(2, 3, 4, 5, 6),``                 ``array``(7, 8, 9, 1, 0),``                 ``array``(9, 6, 4, 2, 3));`` ` `    ``echo` `FindMaxProduct(``\$arr``, ``\$n``);` `// This code is contributed by anuj_67.``?>`

## Javascript

 ``
Output
`3024`

For row-wise adjacent elements, we can generalize the method using a sliding window.

Note: All elements in the matrix must be non-zero.

Another Approach:

1. For each row, create a window of size k. Find the product of k adjacent element as window product (wp).
2. Iterate through the row from k to  (row size), by sliding window approach, find the maximum product. Note: (row size)>=k.
3. Assign the maximum product to a global maximum product.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach``#include ``using` `namespace` `std;` `int` `maxPro(``int` `a, ``int` `n, ``int` `m, ``int` `k)``{``    ``int` `maxi(1), mp(1);``    ``for` `(``int` `i = 0; i < n; ++i)``    ``{``        ``// Window Product for each row.``        ``int` `wp(1);``        ``for` `(``int` `l = 0; l < k; ++l)``        ``{``            ``wp *= a[i][l];``        ``}``      ` `        ``// Maximum window product for each row``        ``mp = wp;``        ``for` `(``int` `j = k; j < m; ++j)``        ``{``            ``wp = wp * a[i][j] / a[i][j - k];``          ` `            ``// Global maximum window product``            ``maxi = max(maxi,max(mp,wp));``        ``}``    ``}``    ``return` `maxi;``}` `// Driver Code``int` `main()``{``    ``int` `n = 6, m = 5, k = 4;``    ``int` `a = { { 1, 2, 3, 4, 5 },``                    ``{ 6, 7, 8, 9, 1 },``                    ``{ 2, 3, 4, 5, 6 },``                    ``{ 7, 8, 9, 1, 0 },``                    ``{ 9, 6, 4, 2, 3 },``                   ``{ 1, 1, 2, 1, 1 } };` `    ``cout << maxPro(a, n, m, k);``    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``import` `java.io.*;` `class` `GFG {``    ``public` `static` `int` `maxPro(``int``[][] a,``                             ``int` `n, ``int` `m,``                             ``int` `k)``    ``{``        ``int` `maxi = ``1``, mp = ``1``;``        ``for` `(``int` `i = ``0``; i < n; ++i)``        ``{``            ``// Window Product for each row.``            ``int` `wp = ``1``;``            ``for` `(``int` `l = ``0``; l < k; ++l)``            ``{``                ``wp *= a[i][l];``            ``}``           ` `            ``// Maximum window product for each row``            ``mp = wp;``            ``for` `(``int` `j = k; j < m; ++j)``            ``{``                ``wp = wp * a[i][j] / a[i][j - k];``              ` `                ``// Global maximum``                ``// window product``                ``maxi = Math.max(``                    ``maxi,``                    ``Math.max(mp, wp));``            ``}``        ``}``        ``return` `maxi;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{` `       ` `        ``int` `n = ``6``, m = ``5``, k = ``4``;``        ``int``[][] a = ``new` `int``[][] {``            ``{ ``1``, ``2``, ``3``, ``4``, ``5` `}, { ``6``, ``7``, ``8``, ``9``, ``1` `},``            ``{ ``2``, ``3``, ``4``, ``5``, ``6` `}, { ``7``, ``8``, ``9``, ``1``, ``0` `},``            ``{ ``9``, ``6``, ``4``, ``2``, ``3` `}, { ``1``, ``1``, ``2``, ``1``, ``1` `}``        ``};``      ` `       ` `        ``// Function call``        ``int` `maxpro = maxPro(a, n, m, k);``        ``System.out.println(maxpro);``    ``}``}`

## Python3

 `# Python implementation of the above approach``def` `maxPro(a,n,m,k):``    ``maxi ``=` `1``    ``mp ``=` `1` `    ``for` `i ``in` `range``(n):``        ` `        ``# Window Product for each row.``        ``wp ``=` `1``        ` `        ``for` `l ``in` `range``(k):``            ``wp ``*``=` `a[i][l]``            ` `        ``# Maximum window product for each row``        ``mp ``=` `wp``        ` `        ``for` `j ``in` `range``(k,m):``            ``wp ``=` `wp ``*` `a[i][j] ``/` `a[i][j ``-` `k]``            ` `            ``# Global maximum``            ``# window product``            ``maxi ``=` `max``(``                    ``maxi,``                    ``max``(mp, wp))``    ` `    ``return` `maxi` `# Driver Code``n ``=` `6``m ``=` `5``k ``=` `4``a``=``[[``1``, ``2``, ``3``, ``4``, ``5` `], [ ``6``, ``7``, ``8``, ``9``, ``1` `],``            ``[ ``2``, ``3``, ``4``, ``5``, ``6` `], [ ``7``, ``8``, ``9``, ``1``, ``0` `],``            ``[ ``9``, ``6``, ``4``, ``2``, ``3` `], [ ``1``, ``1``, ``2``, ``1``, ``1` `]]` `# Function call``maxpro ``=` `maxPro(a, n, m, k)``print``(maxpro)` `# This code is contributed by ab2127`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG{``    ` `public` `static` `int` `maxPro(``int``[,] a, ``int` `n,``                         ``int` `m, ``int` `k)``{``    ``int` `maxi = 1, mp = 1;``    ``for``(``int` `i = 0; i < n; ++i)``    ``{``        ` `        ``// Window Product for each row.``        ``int` `wp = 1;``        ``for``(``int` `l = 0; l < k; ++l)``        ``{``            ``wp *= a[i, l];``        ``}``        ` `        ``// Maximum window product for each row``        ``mp = wp;``        ``for``(``int` `j = k; j < m; ++j)``        ``{``            ``wp = wp * a[i, j] / a[i, j - k];``            ` `            ``// Global maximum``            ``// window product``            ``maxi = Math.Max(maxi,``                   ``Math.Max(mp, wp));``        ``}``    ``}``    ``return` `maxi;``}` `// Driver Code``static` `public` `void` `Main()``{``    ``int` `n = 6, m = 5, k = 4;``    ``int``[,] a = {{ 1, 2, 3, 4, 5 }, { 6, 7, 8, 9, 1 },``                ``{ 2, 3, 4, 5, 6 }, { 7, 8, 9, 1, 0 },``                ``{ 9, 6, 4, 2, 3 }, { 1, 1, 2, 1, 1 }};``   ` `    ``// Function call``    ``int` `maxpro = maxPro(a, n, m, k);``    ``Console.WriteLine(maxpro);``}``}` `// This code is contributed by avanitrachhadiya2155`

## Javascript

 ``
Output
`3024`

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