Maximum sum of any submatrix of a Matrix which is sorted row-wise and column-wise

Given a matrix mat[][] whose elements are sorted both row-wise and column-wise. The task is to find the maximum sum of any submatrix from the given matrix mat[][].
 

Examples:

Input: mat[][] = { {-6, -4, -1}, {-3, 2, 4}, {2, 5, 8}} 
Output: 19 
Explanation: 
The largest submatrix is given by: 
2 4 
5 8
Input: mat[][] = { {-4, -3}, {-2, -1} } 
Output: -1 
Explanation: 
The sub matrix consisting of the last element i.e., -1 has the largest sum possible. 
 

Naive Approach: The idea is to find the maximum sum rectangle in 2D matrix using the Kadane’s Algorithm. Print the maximum sum obtained.

Time Complexity: O(N2*M2), where N is the number of rows and M is the number of columns 
Auxiliary Space: O(N)
Efficient Approach: The idea is to find the maximum sum from bottom cell of the given matrix and use Dynamic Programming to store the maximum sum of any submatrix from bottom cell. Below are the steps:
 



  1. Create a dp table dp[][] of size NxM to store the maximum sum ofsubmatrix starting from each cell (i, j).
  2. Find the sum of the sub-matrix starting from the bottom-right cell (N, M) going up and left and keep updating the maximum sum to dp[][].
  3. Since the matrix is sorted Row-Wise and Column-Wise the largest sub-matrix sum may start from any point, but will definitely end on bottom-right cell (N, M).
  4. Below is the relation how the dp table is filled: 
     

    DP[i][j] = DP[i+1][j] + DP[i][j+1] – DP[i+1][j+1] 
     

  5. Hence, find the maximum element in the dp table.

Below is the implementation of the above approach:
 

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that finds the maximum
// Sub-Matrix Sum
int maxSubMatSum(vector<vector<int> > mat)
{
    // Number of rows in the matrix
    int n = mat.size();
  
    // Number of columns in the matrix
    int m = mat[0].size();
  
    int i, j;
  
    // dp[][] matrix to store the
    // results of each iteration
    int dp[n][m];
  
    // Base Case - The largest
    // element in the matrix
    dp[n - 1][m - 1] = mat[n - 1][m - 1];
  
    // To stores the final result
    int res = dp[n - 1][m - 1];
  
    // Find the max sub matrix sum for
    // the last row
    for (i = m - 2; i >= 0; i--) {
  
        dp[n - 1][i] = mat[n - 1][i]
                       + dp[n - 1][i + 1];
  
        // Check whether the current
        // sub-array yeilds maximum sum
        res = max(res, dp[n - 1][i]);
    }
  
    // Calculate the max sub matrix
    // sum for the last column
    for (i = n - 2; i >= 0; i--) {
  
        dp[i][m - 1] = mat[i][m - 1]
                       + dp[i + 1][m - 1];
  
        // Check whether the current
        // sub-array yeilds maximum sum
        res = max(res, dp[i][m - 1]);
    }
  
    // Build the dp[][] matrix from
    // bottom to the top row
    for (i = n - 2; i >= 0; i--) {
  
        for (j = m - 2; j >= 0; j--) {
  
            // Update sum at each
            // cell in dp[][]
            dp[i][j]
                = mat[i][j] + dp[i][j + 1]
                  + dp[i + 1][j]
                  - dp[i + 1][j + 1];
  
            // Update the maximum sum
            res = max(res, dp[i][j]);
        }
    }
  
    // Return the maximum sum
    return res;
}
  
// Driver Code
int main()
{
    // Given matrix mat[][]
    vector<vector<int> > mat;
    mat = { { -6, -4, -1 },
            { -3, 2, 4 },
            { 2, 5, 8 } };
  
    // Function Call
    cout << maxSubMatSum(mat);
    return 0;
}

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Java

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// Java program for the above approach
class GFG{
  
// Function that finds the maximum
// Sub-Matrix Sum
static int maxSubMatSum(int [][]mat)
{
    // Number of rows in the matrix
    int n = mat.length;
  
    // Number of columns in the matrix
    int m = mat[0].length;
  
    int i, j;
  
    // dp[][] matrix to store the
    // results of each iteration
    int [][]dp = new int[n][m];
  
    // Base Case - The largest
    // element in the matrix
    dp[n - 1][m - 1] = mat[n - 1][m - 1];
  
    // To stores the final result
    int res = dp[n - 1][m - 1];
  
    // Find the max sub matrix sum for
    // the last row
    for (i = m - 2; i >= 0; i--) 
    {
        dp[n - 1][i] = mat[n - 1][i] + 
                        dp[n - 1][i + 1];
  
        // Check whether the current
        // sub-array yeilds maximum sum
        res = Math.max(res, dp[n - 1][i]);
    }
  
    // Calculate the max sub matrix
    // sum for the last column
    for (i = n - 2; i >= 0; i--)
    {
        dp[i][m - 1] = mat[i][m - 1] +
                        dp[i + 1][m - 1];
  
        // Check whether the current
        // sub-array yeilds maximum sum
        res = Math.max(res, dp[i][m - 1]);
    }
  
    // Build the dp[][] matrix from
    // bottom to the top row
    for (i = n - 2; i >= 0; i--) 
    {
        for (j = m - 2; j >= 0; j--)
        {
  
            // Update sum at each
            // cell in dp[][]
            dp[i][j] = mat[i][j] + dp[i][j + 1] + 
                    dp[i + 1][j] - dp[i + 1][j + 1];
  
            // Update the maximum sum
            res = Math.max(res, dp[i][j]);
        }
    }
  
    // Return the maximum sum
    return res;
}
  
// Driver Code
public static void main(String[] args)
{
    // Given matrix mat[][]
    int [][]mat= {{ -6, -4, -1 },
                  { -3, 2, 4 },
                  { 2, 5, 8 } };
  
    // Function Call
    System.out.print(maxSubMatSum(mat));
}
}
  
// This code is contributed by Rohit_ranjan

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Python3

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# Python3 program for the above approach
  
# Function that finds the maximum
# Sub-Matrix Sum
def maxSubMatSum(mat):
      
    # Number of rows in the matrix
    n = len(mat)
  
    # Number of columns in the matrix
    m = len(mat[0])
  
    # dp[][] matrix to store the
    # results of each iteration
    dp = [[0] * m for _ in range(n)]
  
    # Base Case - The largest
    # element in the matrix
    dp[n - 1][m - 1] = mat[n - 1][m - 1]
  
    # To stores the final result
    res = dp[n - 1][m - 1]
  
    # Find the max sub matrix sum for
    # the last row
    for i in range(m - 2, -1, -1):
        dp[n - 1][i] = (mat[n - 1][i] + 
                         dp[n - 1][i + 1])
  
        # Check whether the current
        # sub-array yeilds maximum sum
        res = max(res, dp[n - 1][i])
  
    # Calculate the max sub matrix
    # sum for the last column
    for i in range(n - 2, -1, -1):
        dp[i][m - 1] = (mat[i][m - 1] + 
                     dp[i + 1][m - 1])
  
        # Check whether the current
        # sub-array yeilds maximum sum
        res = max(res, dp[i][m - 1])
  
    # Build the dp[][] matrix from
    # bottom to the top row
    for i in range(n - 2, -1, -1):
        for j in range(m - 2, -1, -1):
  
            # Update sum at each
            # cell in dp[][]
            dp[i][j] = (mat[i][j] + 
                         dp[i][j + 1] +
                         dp[i + 1][j]- 
                         dp[i + 1][j + 1])
  
            # Update the maximum sum
            res = max(res, dp[i][j])
  
    # Return the maximum sum
    return res
  
# Driver Code
if __name__ == '__main__':
      
    # Given matrix mat[][]
    mat = [ [ -6, -4, -1 ],
            [ -3, 2, 4 ],
            [ 2, 5, 8 ] ]
  
    # Function call
    print(maxSubMatSum(mat))
  
# This code is contributed by mohit kumar 29

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C#

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// C# program for the above approach
using System;
class GFG{
  
// Function that finds the maximum
// Sub-Matrix Sum
static int maxSubMatSum(int [,]mat)
{
    // Number of rows in the matrix
    int n = mat.GetLength(0);
  
    // Number of columns in the matrix
    int m = mat.GetLength(1);
  
    int i, j;
  
    // [,]dp matrix to store the
    // results of each iteration
    int [,]dp = new int[n, m];
  
    // Base Case - The largest
    // element in the matrix
    dp[n - 1, m - 1] = mat[n - 1, m - 1];
  
    // To stores the readonly result
    int res = dp[n - 1, m - 1];
  
    // Find the max sub matrix sum for
    // the last row
    for (i = m - 2; i >= 0; i--) 
    {
        dp[n - 1, i] = mat[n - 1, i] + 
                        dp[n - 1, i + 1];
  
        // Check whether the current
        // sub-array yeilds maximum sum
        res = Math.Max(res, dp[n - 1,i]);
    }
  
    // Calculate the max sub matrix
    // sum for the last column
    for (i = n - 2; i >= 0; i--)
    {
        dp[i, m - 1] = mat[i, m - 1] +
                        dp[i + 1, m - 1];
  
        // Check whether the current
        // sub-array yeilds maximum sum
        res = Math.Max(res, dp[i, m - 1]);
    }
  
    // Build the [,]dp matrix from
    // bottom to the top row
    for (i = n - 2; i >= 0; i--) 
    {
        for (j = m - 2; j >= 0; j--)
        {
  
            // Update sum at each
            // cell in [,]dp
            dp[i, j] = mat[i, j] + dp[i, j + 1] + 
                    dp[i + 1, j] - dp[i + 1, j + 1];
  
            // Update the maximum sum
            res = Math.Max(res, dp[i, j]);
        }
    }
  
    // Return the maximum sum
    return res;
}
  
// Driver Code
public static void Main(String[] args)
{
    // Given matrix [,]mat
    int [,]mat= {{ -6, -4, -1 },
                 { -3, 2, 4 },
                 { 2, 5, 8 } };
  
    // Function Call
    Console.Write(maxSubMatSum(mat));
}
}
  
// This code is contributed by Rohit_ranjan

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Output: 

19




 

Time Complexity: O(N*M), where N is the number of rows and M is the number of columns 
Auxiliary Space: O(N*M)
 

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