Maximum product of 4 adjacent elements in matrix

15 Jun, 2022
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
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.
Asked in: Tolexo
Approach: 
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Group 4 elements that are adjacent to each other in each row and calculate their maximum result.Group 4 elements that are adjacent to each other in each column and calculate their maximum results.Group 4 elements that are adjacent to each other in diagonal and calculate their maximum results.Group 4 elements that are adjacent to each other in anti-diagonal and calculate their maximum results.Compare all calculated maximum results.
Group 4 elements that are adjacent to each other in each row and calculate their maximum result.
Group 4 elements that are adjacent to each other in each column and calculate their maximum results.
Group 4 elements that are adjacent to each other in diagonal and calculate their maximum results.
Group 4 elements that are adjacent to each other in anti-diagonal and calculate their maximum results.
Compare all calculated maximum results.
Below is the implementation of the above approach:
C++
Java
Python3
C#
PHP
Javascript
// C++ program to find out the maximum product// in the matrix which four elements are// adjacent to each other in one direction#include <bits/stdc++.h>using namespace std; const int n = 5; // function to find max productint 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 codeint main(){     /* int arr[][4] = {{6, 2, 3, 4},                    {5, 4, 3, 1},                    {7, 4, 5, 6},                    {8, 3, 1, 0}};*/    /* int arr[][5] = {{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[][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}};     cout << FindMaxProduct(arr, n);    return 0;}
// Java program to find out the// maximum product in the matrix// which four elements are adjacent// to each other in one directionclass 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[][4] = {{6, 2, 3, 4},                           {5, 4, 3, 1},                           {7, 4, 5, 6},                           {8, 3, 1, 0}};*/        /* int arr[][5] = {{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 program to find out the maximum# product in the matrix which four elements# are adjacent to each other in one directionn = 5 # function to find max productdef 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 codeif __name__ == "__main__":          # int arr[][4] = {{6, 2, 3, 4},    #                  {5, 4, 3, 1},    #                  {7, 4, 5, 6},    #                  {8, 3, 1, 0}};    # int arr[][5] = {{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# program to find out the// maximum product in the matrix// which four elements are adjacent// to each other in one directionusing 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// PHP 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 productfunction FindMaxProduct( $arr, $n){    $max = 0; $result;     // iterate the rows.    for ( $i = 0; $i < $n; $i++)    {         // iterate the columns.        for ( $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 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.?>
<script> // Javascript program to find out the// maximum product in the matrix// which four elements are adjacent// to each other in one directionlet n = 5;     // function to find max product    function FindMaxProduct(arr,n)    {        let max = 0, result;         // iterate the rows.        for (let i = 0; i < n; i++)        {            // iterate the columns.            for (let 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 arr[][4] = {{6, 2, 3, 4},                        {5, 4, 3, 1},                        {7, 4, 5, 6},                        {8, 3, 1, 0}};*/        /* int arr[][5] = {{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}};*/         let 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 ]];         document.write(FindMaxProduct(arr, n));     // This code is contributed by sravan kumar </script>
3024
Time Complexity: O(n2)
Auxiliary Space: O(1), as no extra space is used
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:
For each row, create a window of size k. Find the product of k adjacent element as window product (wp).Iterate through the row from k to  (row size), by sliding window approach, find the maximum product. Note: (row size)>=k.Assign the maximum product to a global maximum product.
For each row, create a window of size k. Find the product of k adjacent element as window product (wp).
Iterate through the row from k to  (row size), by sliding window approach, find the maximum product. Note: (row size)>=k.
Assign the maximum product to a global maximum product.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std; int maxPro(int a[6][5], 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 Codeint main(){    int n = 6, m = 5, k = 4;    int a[6][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 },                   { 1, 1, 2, 1, 1 } };     cout << maxPro(a, n, m, k);    return 0;}
// Java implementation of the above approachimport 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);    }}
# Python implementation of the above approachdef 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 Coden = 6m = 5k = 4a=[[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 callmaxpro = maxPro(a, n, m, k)print(maxpro) # This code is contributed by ab2127
// C# implementation of the above approachusing 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 Codestatic 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
<script>// Javascript implementation of the above approach         function maxPro(a,n,m,k)    {        let maxi = 1, mp = 1;        for (let i = 0; i < n; ++i)        {            // Window Product for each row.            let wp = 1;            for (let l = 0; l < k; ++l)            {                wp *= a[i][l];            }                         // Maximum window product for each row            mp = wp;            for (let 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    let n = 6, m = 5, k = 4;    let 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    let maxpro = maxPro(a, n, m, k);    document.write(maxpro);         // This code is contributed by rag2127</script>
3024
Time Complexity: O(n2)
Auxiliary Space: O(1), as no extra space is used
shrikanth13
vt_m
ukasp
DivayVijaySingh
tranceispermanent
avanitrachhadiya2155
sravankumar8128
rag2127
ruhelaa48
ab2127
adi1212
Tolexo
Matrix
Technical Scripter
Tolexo
Matrix
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