Maximum sum submatrix

26 Nov, 2021
Prerequisite: Kadane’s algorithm
Given a 2D array arr[][] of dimension N*M, the task is to find the maximum sum sub-matrix from the matrix arr[][].
Examples:
Input: arr[][] = {{0, -2, -7, 0 },  { 9, 2, -6, 2 }, { -4, 1, -4, 1 }, { -1, 8, 0, -2}}Output: 15Explanation: The submatrix {{9, 2}, {-4, 1}, {-1, 8}} has a sum 15, which is the maximum sum possible.
Input: arr[][] = {{1, 2}, {-5, -7}}Output: 3
Naive Approach: The simplest approach is to generate all possible submatrices from the given matrix and calculate their sum. Finally, print the maximum sum obtained.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program for the above approach#include <bits/stdc++.h>using namespace std; // Function to find maximum sum submatrixvoid maxSubmatrixSum(    vector<vector<int> > matrix){    // Stores the number of rows    // and columns in the matrix    int r = matrix.size();    int c = matrix[0].size();     // Stores maximum submatrix sum    int maxSubmatrix = 0;     // Take each row as starting row    for (int i = 0; i < r; i++) {         // Take each column as the        // starting column        for (int j = 0; j < c; j++) {             // Take each row as the            // ending row            for (int k = i; k < r; k++) {                 // Take each column as                // the ending column                for (int l = j; l < c; l++) {                     // Stores the sum of submatrix                    // having topleft index(i, j)                    // and bottom right index (k, l)                    int sumSubmatrix = 0;                     // Iterate the submatrix                    // row-wise and calculate its sum                    for (int m = i; m <= k; m++) {                        for (int n = j; n <= l; n++) {                            sumSubmatrix += matrix[m][n];                        }                    }                     // Update the maximum sum                    maxSubmatrix                        = max(maxSubmatrix,                              sumSubmatrix);                }            }        }    }     // Print the answer    cout << maxSubmatrix;} // Driver Codeint main(){    vector<vector<int> > matrix = { { 0, -2, -7, 0 },                                    { 9, 2, -6, 2 },                                    { -4, 1, -4, 1 },                                    { -1, 8, 0, -2 } };     maxSubmatrixSum(matrix);     return 0;}
// Java program for the above approachimport java.util.*;class GFG{ // Function to find maximum sum submatrixstatic void maxSubmatrixSum(int[][] matrix){       // Stores the number of rows    // and columns in the matrix    int r = matrix.length;    int c = matrix[0].length;     // Stores maximum submatrix sum    int maxSubmatrix = 0;     // Take each row as starting row    for (int i = 0; i < r; i++) {         // Take each column as the        // starting column        for (int j = 0; j < c; j++) {             // Take each row as the            // ending row            for (int k = i; k < r; k++) {                 // Take each column as                // the ending column                for (int l = j; l < c; l++) {                     // Stores the sum of submatrix                    // having topleft index(i, j)                    // and bottom right index (k, l)                    int sumSubmatrix = 0;                     // Iterate the submatrix                    // row-wise and calculate its sum                    for (int m = i; m <= k; m++) {                        for (int n = j; n <= l; n++) {                            sumSubmatrix += matrix[m][n];                        }                    }                     // Update the maximum sum                    maxSubmatrix                        = Math.max(maxSubmatrix,                              sumSubmatrix);                }            }        }    }     // Print the answer    System.out.println(maxSubmatrix);}   // Driver Codepublic static void main(String[] args){    int[][] matrix = { { 0, -2, -7, 0 },                        { 9, 2, -6, 2 },                        { -4, 1, -4, 1 },                        { -1, 8, 0, -2 } };     maxSubmatrixSum(matrix);}} // This code is contributed by susmitakundugoaldanga.
# Python3 program for the above approach # Function to find maximum sum submatrixdef maxSubmatrixSum(matrix):       # Stores the number of rows    # and columns in the matrix    r = len(matrix)    c = len(matrix[0])     # Stores maximum submatrix sum    maxSubmatrix = 0     # Take each row as starting row    for i in range(r):         # Take each column as the        # starting column        for j in range(c):             # Take each row as the            # ending row            for k in range(i, r):                 # Take each column as                # the ending column                for l in range(j, c):                     # Stores the sum of submatrix                    # having topleft index(i, j)                    # and bottom right index (k, l)                    sumSubmatrix = 0                     # Iterate the submatrix                    # row-wise and calculate its sum                    for m in range(i, k + 1):                        for n in range(j, l + 1):                            sumSubmatrix += matrix[m][n]                     # Update the maximum sum                    maxSubmatrix= max(maxSubmatrix, sumSubmatrix)     # Print the answer    print (maxSubmatrix) # Driver Codeif __name__ == '__main__':    matrix = [ [ 0, -2, -7, 0 ],                [ 9, 2, -6, 2 ],                [ -4, 1, -4, 1 ],                [ -1, 8, 0, -2 ] ]     maxSubmatrixSum(matrix)     # This code is contributed by mohit kumar 29.
// C# program to implement// the above approachusing System;public class GFG{// Function to find maximum sum submatrixstatic void maxSubmatrixSum(int[,] matrix){       // Stores the number of rows    // and columns in the matrix    int r = matrix.GetLength(0);    int c = matrix.GetLength(1);     // Stores maximum submatrix sum    int maxSubmatrix = 0;     // Take each row as starting row    for (int i = 0; i < r; i++) {         // Take each column as the        // starting column        for (int j = 0; j < c; j++) {             // Take each row as the            // ending row            for (int k = i; k < r; k++) {                 // Take each column as                // the ending column                for (int l = j; l < c; l++) {                     // Stores the sum of submatrix                    // having topleft index(i, j)                    // and bottom right index (k, l)                    int sumSubmatrix = 0;                     // Iterate the submatrix                    // row-wise and calculate its sum                    for (int m = i; m <= k; m++) {                        for (int n = j; n <= l; n++) {                            sumSubmatrix += matrix[m, n];                        }                    }                     // Update the maximum sum                    maxSubmatrix                        = Math.Max(maxSubmatrix,                              sumSubmatrix);                }            }        }    }     // Print the answer    Console.WriteLine(maxSubmatrix);} // Driver Codepublic static void Main(String []args){    int[,] matrix = { { 0, -2, -7, 0 },                        { 9, 2, -6, 2 },                        { -4, 1, -4, 1 },                        { -1, 8, 0, -2 } };     maxSubmatrixSum(matrix);}} // This code is contributed by sanjoy_62.
<script> // Javascript program for the above approach     // Function to find maximum sum submatrix    function maxSubmatrixSum(matrix)    {         // Stores the number of rows        // and columns in the matrix        var r = matrix.length;        var c = matrix[0].length;         // Stores maximum submatrix sum        var maxSubmatrix = 0;         // Take each row as starting row        for (i = 0; i < r; i++) {             // Take each column as the            // starting column            for (j = 0; j < c; j++) {                 // Take each row as the                // ending row                for (k = i; k < r; k++) {                     // Take each column as                    // the ending column                    for (l = j; l < c; l++) {                         // Stores the sum of submatrix                        // having topleft index(i, j)                        // and bottom right index (k, l)                        var sumSubmatrix = 0;                         // Iterate the submatrix                        // row-wise and calculate its sum                        for (m = i; m <= k; m++) {                            for (n = j; n <= l; n++) {                                sumSubmatrix += matrix[m][n];                            }                        }                         // Update the maximum sum                        maxSubmatrix = Math.max(maxSubmatrix,                        sumSubmatrix);                    }                }            }        }         // Print the answer        document.write(maxSubmatrix);    }     // Driver Code             var matrix = [ [ 0, -2, -7, 0 ],        [ 9, 2, -6, 2 ],        [ -4, 1, -4, 1 ],        [ -1, 8, 0, -2 ] ];         maxSubmatrixSum(matrix); // This code contributed by umadevi9616 </script>
15
 
Time Complexity: O(N6)Auxiliary Space: O(1)
Efficient Approach using Kadane’s Algorithm: The above approach can be optimized using the following observations:
Fix starting and ending column of the required sub-matrix say start and end respectively.
Now, iterate each row and add row sum from starting to ending column to sumSubmatrix and insert this in an array. After iterating each row, perform Kadane’s Algorithm on this newly created array. If the sum obtained by applying Kadane’s algorithm is greater than the overall maximum sum, update the overall maximum sum.
In the above step, the row sum from starting to ending column can be calculated in constant time by creating an auxiliary matrix of size N*M containing the prefix sum of each row.
Follow the steps below to solve the problem:
Initialize a variable, say maxSum as INT_MIN, to store the maximum subarray sum.
Create a matrix prefMatrix[N][M] that stores the prefix array sum of every row of the given matrix.
Traverse the matrix row-wise using i as the row index and j as the column index and perform the following steps:If the value of i is 0, then set prefMatrix[i][j] = A[i][j].Otherwise, set prefMatrix[i][j] = prefMatrix[i][j – 1] + A[i][j].
If the value of i is 0, then set prefMatrix[i][j] = A[i][j].
Otherwise, set prefMatrix[i][j] = prefMatrix[i][j – 1] + A[i][j].
Now for all possible combinations of starting and ending index of the columns of submatrix over the range [0, M] perform the following steps:Initialize an auxiliary array A[] to stores the maximum sum for each row of the current submatrix.Find the sum from starting to ending column using prefMatrix as follows:If the value of start is positive, then store the required sum S as prefMatrix[i][end] – prefMatrix[i][start – 1].Otherwise, update S as prefMatrix[i][end].Insert S in an array arr[].After iterating all rows in the submatrix, perform Kadane’s algorithm on the array A[] and update the maximum sum maxSum as the maximum of maxSum and value obtained by performing the Kadane’s Algorithm in this step.
Initialize an auxiliary array A[] to stores the maximum sum for each row of the current submatrix.
Find the sum from starting to ending column using prefMatrix as follows:If the value of start is positive, then store the required sum S as prefMatrix[i][end] – prefMatrix[i][start – 1].Otherwise, update S as prefMatrix[i][end].
If the value of start is positive, then store the required sum S as prefMatrix[i][end] – prefMatrix[i][start – 1].
Otherwise, update S as prefMatrix[i][end].
Insert S in an array arr[].
After iterating all rows in the submatrix, perform Kadane’s algorithm on the array A[] and update the maximum sum maxSum as the maximum of maxSum and value obtained by performing the Kadane’s Algorithm in this step.
After completing the above steps, print the value of maxSum as the result.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program for the above approach#include <bits/stdc++.h>using namespace std; // Function to find maximum continuous// maximum sum in the arrayint kadane(vector<int> v){       // Stores current and maximum sum    int currSum = 0;    int maxSum = INT_MIN;     // Traverse the array v    for (int i = 0;         i < (int)v.size(); i++) {         // Add the value of the        // current element        currSum += v[i];         // Update the maximum sum        if (currSum > maxSum) {            maxSum = currSum;        }         if (currSum < 0) {            currSum = 0;        }    }     // Return the maximum sum    return maxSum;} // Function to find the maximum// submatrix sumvoid maxSubmatrixSum(    vector<vector<int> > A){       // Store the rows and columns    // of the matrix    int r = A.size();    int c = A[0].size();     // Create an auxiliary matrix    int** prefix = new int*[r];     // Traverse the matrix, prefix    // and initialize it will all 0s    for (int i = 0; i < r; i++) {         prefix[i] = new int;        for (int j = 0; j < c; j++) {            prefix[i][j] = 0;        }    }     // Calculate prefix sum of all    // rows of matrix A[][] and    // store in matrix prefix[]    for (int i = 0; i < r; i++) {         for (int j = 0; j < c; j++) {             // Update the prefix[][]            if (j == 0)                prefix[i][j] = A[i][j];            else                prefix[i][j] = A[i][j]                               + prefix[i][j - 1];        }    }     // Store the maximum submatrix sum    int maxSum = INT_MIN;     // Iterate for starting column    for (int i = 0; i < c; i++) {         // Iterate for last column        for (int j = i; j < c; j++) {             // To store current array            // elements            vector<int> v;             // Traverse every row            for (int k = 0; k < r; k++) {                 // Store the sum of the                // kth row                int el = 0;                 // Update the prefix                // sum                if (i == 0)                    el = prefix[k][j];                else                    el = prefix[k][j]                         - prefix[k][i - 1];                 // Push it in a vector                v.push_back(el);            }             // Update the maximum            // overall sum            maxSum = max(maxSum, kadane(v));        }    }     // Print the answer    cout << maxSum << "\n";} // Driver Codeint main(){    vector<vector<int> > matrix = { { 0, -2, -7, 0 },                                    { 9, 2, -6, 2 },                                    { -4, 1, -4, 1 },                                    { -1, 8, 0, -2 } };     // Function Call    maxSubmatrixSum(matrix);     return 0;}
// Java program for the above approachimport java.util.*;class GFG{   // Function to find maximum continuous  // maximum sum in the array  static int kadane(Vector<Integer> v)  {     // Stores current and maximum sum    int currSum = 0;    int maxSum = Integer.MIN_VALUE;     // Traverse the array v    for (int i = 0;         i < (int)v.size(); i++)    {       // Add the value of the      // current element      currSum += v.get(i);       // Update the maximum sum      if (currSum > maxSum)      {        maxSum = currSum;      }       if (currSum < 0)      {        currSum = 0;      }    }     // Return the maximum sum    return maxSum;  }   // Function to find the maximum  // submatrix sum  static void maxSubmatrixSum(int [][]A)  {    // Store the rows and columns    // of the matrix    int r = A.length;    int c = A[0].length;     // Create an auxiliary matrix    int [][]prefix = new int[r][];     // Traverse the matrix, prefix    // and initialize it will all 0s    for (int i = 0; i < r; i++) {       prefix[i] = new int;      for (int j = 0; j < c; j++) {        prefix[i][j] = 0;      }    }     // Calculate prefix sum of all    // rows of matrix A[][] and    // store in matrix prefix[]    for (int i = 0; i < r; i++) {       for (int j = 0; j < c; j++) {         // Update the prefix[][]        if (j == 0)          prefix[i][j] = A[i][j];        else          prefix[i][j] = A[i][j]          + prefix[i][j - 1];      }    }     // Store the maximum submatrix sum    int maxSum = Integer.MIN_VALUE;     // Iterate for starting column    for (int i = 0; i < c; i++) {       // Iterate for last column      for (int j = i; j < c; j++) {         // To store current array        // elements        Vector<Integer> v = new Vector<Integer>();         // Traverse every row        for (int k = 0; k < r; k++) {           // Store the sum of the          // kth row          int el = 0;           // Update the prefix          // sum          if (i == 0)            el = prefix[k][j];          else            el = prefix[k][j]            - prefix[k][i - 1];           // Push it in a vector          v.add(el);        }         // Update the maximum        // overall sum        maxSum = Math.max(maxSum, kadane(v));      }    }     // Print the answer    System.out.print(maxSum+ "\n");  }   // Driver Code  public static void main(String[] args)  {    int [][]matrix = { { 0, -2, -7, 0 },                      { 9, 2, -6, 2 },                      { -4, 1, -4, 1 },                      { -1, 8, 0, -2 } };     // Function Call    maxSubmatrixSum(matrix);  }} // This code is contributed by 29AjayKumar
# Python3 program for the above approachimport sys # Function to find maximum continuous# maximum sum in the arraydef kadane(v):         # Stores current and maximum sum    currSum = 0         maxSum = -sys.maxsize - 1         # Traverse the array v    for i in range(len(v)):                 # Add the value of the        # current element        currSum += v[i]                 # Update the maximum sum        if (currSum > maxSum):            maxSum = currSum        if (currSum < 0):            currSum = 0         # Return the maximum sum    return maxSum # Function to find the maximum# submatrix sumdef maxSubmatrixSum(A):         # Store the rows and columns    # of the matrix    r = len(A)    c = len(A[0])         # Create an auxiliary matrix    # Traverse the matrix, prefix    # and initialize it will all 0s    prefix = [[0 for i in range(c)]                 for j in range(r)]         # Calculate prefix sum of all    # rows of matrix A[][] and    # store in matrix prefix[]    for i in range(r):        for j in range(c):                         # Update the prefix[][]            if (j == 0):                prefix[i][j] = A[i][j]            else:                prefix[i][j] = A[i][j] + prefix[i][j - 1]         # Store the maximum submatrix sum    maxSum = -sys.maxsize - 1         #  Iterate for starting column    for i in range(c):                 # Iterate for last column        for j in range(i, c):                         # To store current array            # elements            v = []                         # Traverse every row            for k in range(r):                                 # Store the sum of the                # kth row                el = 0                                 # Update the prefix                # sum                if (i == 0):                    el = prefix[k][j]                else:                    el = prefix[k][j] - prefix[k][i - 1]                                 # Push it in a vector                v.append(el)                         # Update the maximum            # overall sum            maxSum = max(maxSum, kadane(v))         # Print the answer    print(maxSum) # Driver Codematrix = [ [ 0, -2, -7, 0 ],           [ 9, 2, -6, 2 ],           [ -4, 1, -4, 1 ],           [ -1, 8, 0, -2 ] ]            # Function CallmaxSubmatrixSum(matrix) # This code is contributed by rag2127
// C# program for the above approachusing System;using System.Collections.Generic; public class GFG{   // Function to find maximum continuous  // maximum sum in the array  static int kadane(List<int> v)  {     // Stores current and maximum sum    int currSum = 0;    int maxSum = int.MinValue;     // Traverse the array v    for (int i = 0;         i < (int)v.Count; i++)    {       // Add the value of the      // current element      currSum += v[i];       // Update the maximum sum      if (currSum > maxSum)      {        maxSum = currSum;      }       if (currSum < 0)      {        currSum = 0;      }    }     // Return the maximum sum    return maxSum;  }   // Function to find the maximum  // submatrix sum  static void maxSubmatrixSum(int [,]A)  {    // Store the rows and columns    // of the matrix    int r = A.GetLength(0);    int c = A.GetLength(1);     // Create an auxiliary matrix    int [,]prefix = new int[r,c];     // Traverse the matrix, prefix    // and initialize it will all 0s    for (int i = 0; i < r; i++) {      for (int j = 0; j < c; j++) {        prefix[i,j] = 0;      }    }     // Calculate prefix sum of all    // rows of matrix [,]A and    // store in matrix prefix[]    for (int i = 0; i < r; i++) {       for (int j = 0; j < c; j++) {         // Update the prefix[,]        if (j == 0)          prefix[i,j] = A[i,j];        else          prefix[i,j] = A[i,j]          + prefix[i,j - 1];      }    }     // Store the maximum submatrix sum    int maxSum = int.MinValue;     // Iterate for starting column    for (int i = 0; i < c; i++) {       // Iterate for last column      for (int j = i; j < c; j++) {         // To store current array        // elements        List<int> v = new List<int>();         // Traverse every row        for (int k = 0; k < r; k++) {           // Store the sum of the          // kth row          int el = 0;           // Update the prefix          // sum          if (i == 0)            el = prefix[k,j];          else            el = prefix[k,j]            - prefix[k,i - 1];           // Push it in a vector          v.Add(el);        }         // Update the maximum        // overall sum        maxSum = Math.Max(maxSum, kadane(v));      }    }     // Print the answer    Console.Write(maxSum+ "\n");  }   // Driver Code  public static void Main(String[] args)  {    int [,]matrix = { { 0, -2, -7, 0 },                      { 9, 2, -6, 2 },                      { -4, 1, -4, 1 },                      { -1, 8, 0, -2 } };     // Function Call    maxSubmatrixSum(matrix);  }} // This code is contributed by 29AjayKumar
<script>// Javascript program for the above approach // Function to find maximum continuous// maximum sum in the array   function kadane(v){    // Stores current and maximum sum    let currSum = 0;    let maxSum = Number.MIN_VALUE;      // Traverse the array v    for (let i = 0;         i < v.length; i++)    {        // Add the value of the      // current element      currSum += v[i];        // Update the maximum sum      if (currSum > maxSum)      {        maxSum = currSum;      }        if (currSum < 0)      {        currSum = 0;      }    }      // Return the maximum sum    return maxSum;} // Function to find the maximum// submatrix sumfunction maxSubmatrixSum(A){    // Store the rows and columns    // of the matrix    let r = A.length;    let c = A[0].length;      // Create an auxiliary matrix    let prefix = new Array(r);      // Traverse the matrix, prefix    // and initialize it will all 0s    for (let i = 0; i < r; i++) {        prefix[i] = new Array(c);      for (let j = 0; j < c; j++) {        prefix[i][j] = 0;      }    }      // Calculate prefix sum of all    // rows of matrix A[][] and    // store in matrix prefix[]    for (let i = 0; i < r; i++) {        for (let j = 0; j < c; j++) {          // Update the prefix[][]        if (j == 0)          prefix[i][j] = A[i][j];        else          prefix[i][j] = A[i][j]          + prefix[i][j - 1];      }    }      // Store the maximum submatrix sum    let maxSum = Number.MIN_VALUE;      // Iterate for starting column    for (let i = 0; i < c; i++) {        // Iterate for last column      for (let j = i; j < c; j++) {          // To store current array        // elements        let v = [];          // Traverse every row        for (let k = 0; k < r; k++) {            // Store the sum of the          // kth row          let el = 0;            // Update the prefix          // sum          if (i == 0)            el = prefix[k][j];          else            el = prefix[k][j]            - prefix[k][i - 1];            // Push it in a vector          v.push(el);        }          // Update the maximum        // overall sum        maxSum = Math.max(maxSum, kadane(v));      }    }      // Print the answer    document.write(maxSum+ "<br>");} // Driver Codelet matrix=[[ 0, -2, -7, 0 ],                      [ 9, 2, -6, 2 ],                      [ -4, 1, -4, 1 ],                      [ -1, 8, 0, -2 ]];// Function CallmaxSubmatrixSum(matrix); // This code is contributed by unknown2108</script>
15
 
Time Complexity: O(N3)Auxiliary Space: O(N2)
mohit kumar 29
susmitakundugoaldanga
sanjoy_62
29AjayKumar
rag2127
umadevi9616
unknown2108
ankita_saini
prefix-sum
subarray
subarray-sum
submatrix
Technical Scripter 2020
Dynamic Programming
Mathematical
Matrix
Technical Scripter
prefix-sum
Dynamic Programming
Mathematical
Matrix
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