Maximum sum submatrix - GeeksforGeeks (2024)

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C++ Java Python3 C# Javascript C++ Java Python3 C# Javascript Please Login to comment...

Last Updated : 11 Jul, 2022

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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: 15
Explanation: 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++

// C++ program for the above approach

#include <bits/stdc++.h>

using namespace std;

// Function to find maximum sum submatrix

void 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 Code

int 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
 
// Java program for the above approach 
import java.util.*; 
class GFG 
{ 
// Function to find maximum sum submatrix 
static 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 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 } }; 
maxSubmatrixSum(matrix); 
} 
} 
// This code is contributed by susmitakundugoaldanga.
 
 
Python3
 
# Python3 program for the above approach 
# Function to find maximum sum submatrix 
def 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 Code 
if __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#
 
// C# program to implement 
// the above approach 
using System; 
public class GFG 
{ 
// Function to find maximum sum submatrix 
static 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; 
See Also
Sum of all Submatrices of a Given Matrix - GeeksforGeeksSolution: Number of Submatrices That Sum to TargetNumber of Submatrices That Sum to TargetSubmatrix Sum Queries - GeeksforGeeks
// 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 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 } }; 
maxSubmatrixSum(matrix); 
} 
} 
// This code is contributed by sanjoy_62.
 
 
Javascript
 
<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> 
 
 
Output:
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].
  • 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.
  • After completing the above steps, print the value of maxSum as the result.
Below is the implementation of the above approach:
C++
 
// C++ program for the above approach 
#include <bits/stdc++.h> 
using namespace std; 
// Function to find maximum continuous 
// maximum sum in the array 
int 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 sum 
void 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 Code 
int 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
 
// Java program for the above approach 
import 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
 
# Python3 program for the above approach 
import sys 
# Function to find maximum continuous 
# maximum sum in the array 
def 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 sum 
def 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 Code 
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 rag2127
 
 
C#
 
// C# program for the above approach 
using 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
 
 
Javascript
 
<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 sum 
function 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 Code 
let 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 unknown2108 
</script> 
 
 
Output:
15
Time Complexity: O(N3)
Auxiliary Space: O(N2)
Related Topic: Subarrays, Subsequences, and Subsets in Array

 
 
 
 
 
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