# Find the maximum sum of Plus shape pattern in a 2-D array

• Difficulty Level : Medium
• Last Updated : 12 Sep, 2022

Given a 2-D array of size N*M where, . The task is to find the maximum value achievable by a + shaped pattern. The elements of the array can be negative.
The plus(+) shape pattern is formed by taking any element with co-ordinate (x, y) as a center and then expanding it in all four directions(if possible)

A plus(+) shape has atleast five elements which are { (x-1, y), (x, y-1), (x, y), (x+1, y), (x, y+1) } i.e. the arms should have length>1 but not necessarily need to have same length.

Examples:

Input: N = 3, M = 4
1 1 1 1
-6 1 1 -4
1 1 1 1
Output: 0
Here, (x, y)=(2, 3) center of pattern(+).
Other four arms are, left arm = (2, 2), right arm = (2, 4),
up arm = (1, 3), down arm = (2, 3).
Hence sum of all elements are ( 1 + 1 + (-4) + 1 + 1 ) = 0.

Input: N = 5, M = 3
1 2 3
-6 1 -4
1 1 1
7 8 9
6 3 2
Output: 31

Approach: This problem is an application of the standard Largest Sum Contiguous Subarray.

We quickly pre-compute the maximum contiguous sub-sequence (subarray) sum for each row and column, in 4 directions, namely, Up, Down, Left and Right. This can be done using the standard Maximum contiguous sub-sequence sum of a 1-D array.

We make four 2-D array’s 1 for each direction.

1. up[i][j]– Maximum sum contiguous sub-sequence of elements in upward direction, from rows 1, 2, 3, …, i More formally, it represents the maximum sum obtained by adding a contiguous sub-sequence of elements from list of arr[j], arr[j], …, arr[i][j]
2. down[i][j] -Maximum sum contiguous sub-sequence of elements in downward direction, from rows i, i+1, i+2,,…, N More formally, it represents the maximum sum obtained by adding a contiguous sub-sequence of elements from list of arr[i][j], arr[i+1][j], …, arr[N][j]
3. left[i][j]– Maximum sum contiguous sub-sequence of elements in left direction, from columns 1, 2, 3, …, j More formally, it represents the maximum sum obtained by adding a contiguous sub-sequence of elements from list of arr[i], arr[i], …, arr[i][j]
4. right[i][j]– Maximum sum contiguous sub-sequence of elements in right direction, from columns j, j+1, j+2, …, M More formally, it represents the maximum sum obtained by adding a contiguous sub-sequence of elements from list of arr[i][j], arr[i][j+1], …, arr[i][M]

All that’s left is, to check each cell as a possible center of the + and use pre-computed data to find the value achieved by + shape in O(1). Below is the implementation of the above approach:

## C++

 // C++ program to find the maximum value// of a + shaped pattern in 2-D array#include using namespace std;#define N 100 const int n = 3, m = 4; // Function to return maximum Plus valueint maxPlus(int (&arr)[n][m]){     // Initializing answer with the minimum value    int ans = INT_MIN;     // Initializing all four arrays    int left[N][N], right[N][N], up[N][N], down[N][N];     // Initializing left and up array.    for (int i = 0; i < n; i++) {        for (int j = 0; j < m; j++) {            left[i][j] = max(0LL, (j ? left[i][j - 1] : 0LL))                                              + arr[i][j];            up[i][j] = max(0LL, (i ? up[i - 1][j] : 0LL))                                              + arr[i][j];        }    }     // Initializing right and down array.    for (int i = 0; i < n; i++) {        for (int j = 0; j < m; j++) {            right[i][j] = max(0LL, (j + 1 == m ? 0LL: right[i][j + 1]))                                                            + arr[i][j];            down[i][j] = max(0LL, (i + 1 == n ? 0LL: down[i + 1][j]))                                                            + arr[i][j];        }    }     // calculating value of maximum Plus (+) sign    for (int i = 1; i < n - 1; ++i)        for (int j = 1; j < m - 1; ++j)            ans = max(ans, up[i - 1][j] + down[i + 1][j]                        + left[i][j - 1] + right[i][j + 1] + arr[i][j]);     return ans;} // Driver codeint main(){     int arr[n][m] = { { 1, 1, 1, 1 },                      { -6, 1, 1, -4 },                      { 1, 1, 1, 1 } };     // Function call to find maximum value    cout << maxPlus(arr);     return 0;}

## Java

 // Java program to find the maximum value// of a + shaped pattern in 2-D array     class GFG{    public static int N = 100;         public static int n = 3, m = 4;             // Function to return maximum Plus value    public static int maxPlus(int[][] arr)    {                 // Initializing answer with the minimum value        int ans = Integer.MIN_VALUE;                 // Initializing all four arrays        int[][] left = new int[N][N];        int[][] right = new int[N][N];        int[][] up = new int[N][N];        int[][] down = new int[N][N];                 // Initializing left and up array.        for (int i = 0; i < n; i++)        {            for (int j = 0; j < m; j++)            {                left[i][j] = Math.max(0, ((j != 0) ? left[i][j - 1] : 0))                                                + arr[i][j];                up[i][j] = Math.max(0, ((i != 0)? up[i - 1][j] : 0))                                                + arr[i][j];            }        }                 // Initializing right and down array.        for (int i = 0; i < n; i++)        {            for (int j = 0; j < m; j++)            {                right[i][j] = Math.max(0, (j + 1 == m ? 0: right[i][j + 1]))                                                                + arr[i][j];                down[i][j] = Math.max(0, (i + 1 == n ? 0: down[i + 1][j]))                                                                + arr[i][j];            }        }                 // calculating value of maximum Plus (+) sign        for (int i = 1; i < n - 1; ++i)            for (int j = 1; j < m - 1; ++j)                ans = Math.max(ans, up[i - 1][j] + down[i + 1][j]                            + left[i][j - 1] + right[i][j + 1] + arr[i][j]);                 return ans;    }             // Driver code    public static void main(String[] args) {        int[][] arr = new int[][]{ { 1, 1, 1, 1 },                                   { -6, 1, 1, -4 },                                   { 1, 1, 1, 1 } };        // Function call to find maximum value        System.out.println( maxPlus(arr) );    }} // This code is contributed by PrinciRaj1992.

## Python 3

 # Python 3 program to find the maximum value# of a + shaped pattern in 2-D array N = 100 n = 3m = 4 # Function to return maximum# Plus valuedef maxPlus(arr):     # Initializing answer with    # the minimum value    ans = 0     # Initializing all four arrays    left = [[0 for x in range(N)]               for y in range(N)]    right = [[0 for x in range(N)]                for y in range(N)]    up = [[0 for x in range(N)]             for y in range(N)]    down = [[0 for x in range(N)]               for y in range(N)]     # Initializing left and up array.    for i in range(n) :        for j in range(m) :            left[i][j] = (max(0, (left[i][j - 1] if j else 0)) +                                  arr[i][j])            up[i][j] = (max(0, (up[i - 1][j] if i else 0)) +                                arr[i][j])      # Initializing right and down array.    for i in range(n) :        for j in range(m) :            right[i][j] = max(0, (0 if (j + 1 == m ) else                                  right[i][j + 1])) + arr[i][j]            down[i][j] = max(0, (0 if (i + 1 == n ) else                                 down[i + 1][j])) + arr[i][j]     # calculating value of maximum    # Plus (+) sign    for i in range(1, n - 1):        for j in range(1, m - 1):            ans = max(ans, up[i - 1][j] + down[i + 1][j] +                         left[i][j - 1] + right[i][j + 1] +                         arr[i][j])     return ans # Driver codeif __name__ == "__main__":    arr = [[ 1, 1, 1, 1 ],        [ -6, 1, 1, -4 ],        [ 1, 1, 1, 1 ]]     # Function call to find maximum value    print(maxPlus(arr)) # This code is contributed# by ChitraNayal

## C#

 // C# program to find the maximum value// of a + shaped pattern in 2-D arrayusing System;   class GFG{    public static int N = 100;       public static int n = 3, m = 4;           // Function to return maximum Plus value    public static int maxPlus(int[,] arr)    {               // Initializing answer with the minimum value        int ans = int.MinValue;               // Initializing all four arrays        int[,] left = new int[N,N];        int[,] right = new int[N,N];        int[,] up = new int[N,N];        int[,] down = new int[N,N];               // Initializing left and up array.        for (int i = 0; i < n; i++) {            for (int j = 0; j < m; j++) {                left[i,j] = Math.Max(0, ((j != 0) ? left[i,j - 1] : 0))                                                   + arr[i,j];                up[i,j] = Math.Max(0, ((i != 0)? up[i - 1,j] : 0))                                                  + arr[i,j];            }        }               // Initializing right and down array.        for (int i = 0; i < n; i++) {            for (int j = 0; j < m; j++) {                right[i,j] = Math.Max(0, (j + 1 == m ? 0: right[i,j + 1]))                                                                + arr[i,j];                down[i,j] = Math.Max(0, (i + 1 == n ? 0: down[i + 1,j]))                                                                + arr[i,j];            }        }               // calculating value of maximum Plus (+) sign        for (int i = 1; i < n - 1; ++i)            for (int j = 1; j < m - 1; ++j)                ans = Math.Max(ans, up[i - 1,j] + down[i + 1,j]                             + left[i,j - 1] + right[i,j + 1] + arr[i,j]);               return ans;    }           // Driver code    static void Main()    {        int[,] arr = new int[,]{ { 1, 1, 1, 1 },                      { -6, 1, 1, -4 },                      { 1, 1, 1, 1 } };           // Function call to find maximum value        Console.Write( maxPlus(arr) );    }} // This code is contributed by DrRoot_

## Javascript

 

Output

0

Time Complexity: O(N*M) for given N rows and M columns

Auxiliary Space: O(N*M)

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