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 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 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# 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;
// 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.
<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>
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[c];
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 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[c];
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 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# 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
<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>
Time Complexity: O(N3)
Auxiliary Space: O(N2)
Related Topic: Subarrays, Subsequences, and Subsets in Array
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Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
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Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
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Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
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Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
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Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
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GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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