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

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[1][j], arr[2][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][1], arr[i][2], …, 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 <bits/stdc++.h>
using namespace std;
#define N 100

const int n = 3, m = 4;

// Function to return maximum Plus value
int 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 code
int 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 = 3
m = 4

# Function to return maximum
# Plus value
def 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 code
if __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 array
using 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

<script>

// JavaScript program to find the maximum value
// of a + shaped pattern in 2-D array

    let N = 100;
    let n = 3, m = 4;

    //Function to return maximum Plus value

    function maxPlus(arr)
    {
        // Initializing answer with the minimum value
        let ans = 0;

        // Initializing all four arrays
        let left = new Array(N);
        let right = new Array(N);
        let up = new Array(N);
        let down = new Array(N);
        for(let i=0;i<N;i++)
        {
            left[i]=new Array(N);
            right[i]=new Array(N);
            up[i]=new Array(N);
            down[i]=new Array(N);
            for(let j=0;j<N;j++)
            {
                left[i][j]=0;
                right[i][j]=0;
                up[i][j]=0;
                down[i][j]=0;
            }

        }

        // Initializing left and up array.
        for (let i = 0; i < n; i++)
        {
            for (let 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 (let i = 0; i < n; i++)
        {
            for (let 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 (let i = 1; i < n - 1; ++i)
            for (let 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
    let arr = [[ 1, 1, 1, 1 ],
        [ -6, 1, 1, -4 ],
        [ 1, 1, 1, 1 ]];
    document.write(maxPlus(arr));



// This code is contributed by avanitrachhadiya2155

</script>

Output

0

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

Auxiliary Space: O(N*M)