Related Articles
Maximum product of 4 adjacent elements in matrix
• Difficulty Level : Easy
• Last Updated : 20 Jan, 2021

Given a square matrix, find the maximum product of four adjacent elements of matrix. The adjacent elements of 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

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

Below is the implementation of 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.` `?>`

Output

`3024`

For row-wise adjacent elements, we can generalize the method using 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++ implemenation 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);` `    ``}` `}`

Output

`3024`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :