Given a 2D array, find the maximum sum submatrix in it. For example, in the following 2D array, the maximum sum submatrix is highlighted with blue rectangle and sum of all elements in this submatrix is 29.
This problem is mainly an extension of Largest Sum Contiguous Subarray for 1D array.
The Naive Solution for this problem is to check every possible rectangle in the given 2D array. This solution requires 6 nested loops –
- 4 for start and end coordinate of the 2 axis O(n4)
- and 2 for the summation of the sub-matrix O(n2).
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>using namespace std;void maxMatrixSum(vector<vector<int> >& matrix){ int n = matrix.size(); // no of rows in a matrix; int m = matrix[0].size(); // no of columns in matrix; int maxsum = INT_MIN; int top = 0, bottom = 0, left = 0, right = 0; // a loop for top row position in the // rectangle for (int i = 0; i < n; i++) { // a loop for left column position // of the rectangle for (int j = 0; j < m; j++) { // a loop for bottom row in the // rectangle for (int k = 0; k < n; k++) { for (int l = 0; l < m; l++) { // a loop for right column in // the rectangle int curr = 0; for (int x = i; x <= k; x++) { // a loops execute for // finding sum of elements in the // rectangle for (int y = j; y <= l; y++) { // for all possibble // points of rectangle curr += matrix[x][y]; } } // updating the resultant variables if curr > maxsum if (curr > maxsum) { maxsum = curr; top = i; left = j; right = k; bottom = l; } } } } } cout << "( Top , Left )" << "( " << top << " , " << left << " )" << endl; cout << "( Bottom , Right )" << "( " << bottom << " , " << right << " )" << endl; cout << "The sum of this rectangle is: " << maxsum << endl;}int main(){ vector<vector<int> > v = { { 1, 2, -1, -4, -20 }, { -8, -3, 4, 2, 1 }, { 3, 8, 10, 1, 3 }, { -4, -1, 1, 7, -6 } }; maxMatrixSum(v); return 0;}// contributed by hungry_coder_109(Naveen); |
Java
import java.io.*;import java.lang.*;import java.util.*;class GFG { public static void maxMatrixSum(int[][] matrix) { int n = matrix.length; // no of rows in a matrix; int m = matrix[0].length; // no of columns in matrix; int maxsum = -999999999; int top = 0, bottom = 0, left = 0, right = 0; for (int i = 0; i < n; i++) { // This loop for top row position in the // rectangle for (int j = 0; j < m; j++) { // This loop for left column // position of the rectangle for (int k = 0; k < n; k++) { // This loop for bottom row in // the rectangle for (int l = 0; l < m; l++) { // This loop for right // column in the rectangle int curr = 0; for (int x = i; x <= k; x++) { // This loops execute // for finding sum of // elements in the // rectangle for (int y = j; y <= l; y++) { // for all possibble // points of // rectangle curr += matrix[x][y]; } } if (curr > maxsum) { maxsum = curr; top = i; left = j; right = k; bottom = l; } } } } } System.out.println("Top , Left ) ( " + top + " , " + left + " )"); System.out.println("Bottom , Right) ( " + bottom + " , " + right + " )"); System.out.println("The sum of the rectangle is: " + maxsum); } public static void main(String[] args) { int arr[][] = new int[][] { { 1, 2, -1, -4, -20 }, { -8, -3, 4, 2, 1 }, { 3, 8, 10, 1, 3 }, { -4, -1, 1, 7, -6 } }; // Function call maxMatrixSum(arr); }}// contributed by hungry_coder_109(Naveen); |
Python3
def max_matrix_sum(matrix): n = len(matrix) # number of rows in the matrix m = len(matrix[0]) # number of columns in the matrix max_sum = float('-inf') top = 0 bottom = 0 left = 0 right = 0 # Loop for top row position in the rectangle for i in range(n): # Loop for left column position of the rectangle for j in range(m): # Loop for bottom row in the rectangle for k in range(n): # Loop for right column in the rectangle for l in range(m): curr = 0 # Loop to find the sum of elements in the rectangle for x in range(i, k+1): # Loop for all possible points of the rectangle for y in range(j, l+1): curr += matrix[x][y] # Update the result variables if curr > max_sum if curr > max_sum: max_sum = curr top = i left = j right = l bottom = k print("(Top, Left): ({}, {})".format(top, left)) print("(Bottom, Right): ({}, {})".format(bottom, right)) print("The sum of this rectangle is:", max_sum)matrix = [[1, 2, -1, -4, -20], [-8, -3, 4, 2, 1], [3, 8, 10, 1, 3], [-4, -1, 1, 7, -6]]max_matrix_sum(matrix) |
C#
using System;class GFG { // Function to find the maximum sum subrectangle in a // given matrix static void MaxMatrixSum(int[][] matrix) { int n = matrix.Length; // Number of rows in the matrix int m = matrix[0].Length; // Number of columns in // the matrix int maxsum = int.MinValue; // Variable to store the maximum // sum found so far int top = 0, bottom = 0, left = 0, right = 0; // Variables to store the coordinates of // the subrectangle with the maximum sum // Loop for the top row position of the rectangle for (int i = 0; i < n; i++) { // Loop for the left column position of the // rectangle for (int j = 0; j < m; j++) { // Loop for the bottom row position of the // rectangle for (int k = i; k < n; k++) { // Loop for the right column position of // the rectangle for (int l = j; l < m; l++) { int curr = 0; // Variable to store the // current sum of elements // in the rectangle // Nested loops execute for finding // the sum of elements in the // rectangle for (int x = i; x <= k; x++) { for (int y = j; y <= l; y++) { curr += matrix[x][y]; } } // Updating the resultant variables // if curr > maxsum if (curr > maxsum) { maxsum = curr; top = i; left = j; right = l; bottom = k; } } } } } // Printing the coordinates and sum of the maximum // sum subrectangle Console.WriteLine("(Top, Left) (" + top + ", " + left + ")"); Console.WriteLine("(Bottom, Right) (" + bottom + ", " + right + ")"); Console.WriteLine("The sum of this rectangle is: " + maxsum); } static void Main() { int[][] v = new int[][] { new int[] { 1, 2, -1, -4, -20 }, new int[] { -8, -3, 4, 2, 1 }, new int[] { 3, 8, 10, 1, 3 }, new int[] { -4, -1, 1, 7, -6 } }; MaxMatrixSum(v); }} |
Javascript
function maxMatrixSum(matrix) { const n = matrix.length; const m = matrix[0].length; let maxsum = Number.MIN_SAFE_INTEGER; let top = 0, bottom = 0, left = 0, right = 0; // A loop for the top row position in // the rectangle for (let i = 0; i < n; i++) { // A loop for the left column // position of the rectangle for (let j = 0; j < m; j++) { for (let k = i; k < n; k++) { for (let l = j; l < m; l++) { let curr = 0; // Loop to calculate the sum of // elements in the rectangle for (let x = i; x <= k; x++) { for (let y = j; y <= l; y++) { curr += matrix[x][y]; } } // Update the result variables if curr > maxsum if (curr > maxsum) { maxsum = curr; top = i; left = j; bottom = k; right = l; } } } } } console.log(`( Top , Left ) ( ${top} , ${left} )`); console.log(`( Bottom , Right ) ( ${bottom} , ${right} )`); console.log(`The sum of this rectangle is: ${maxsum}`);}const matrix = [ [1, 2, -1, -4, -20], [-8, -3, 4, 2, 1], [3, 8, 10, 1, 3], [-4, -1, 1, 7, -6]];maxMatrixSum(matrix); |
Output:
( Top , Left )( 1 , 1 )
( Bottom , Right )( 3 , 3 )
The sum of this rectangle is: 29
Time Complexity: O(n^3 * m^3) where n is the number of rows and m is the numbr of columns in the given matrix.
Efficient Approach –
Kadane’s algorithm for 1D array can be used to reduce the time complexity to O(n^3). The idea is to fix the left and right columns one by one and find the maximum sum contiguous rows for every left and right column pair. We basically find top and bottom row numbers (which have maximum sum) for every fixed left and right column pair. To find the top and bottom row numbers, calculate the sum of elements in every row from left to right and store these sums in an array say temp[]. So temp[i] indicates sum of elements from left to right in row i. If we apply Kadane’s 1D algorithm on temp[], and get the maximum sum subarray of temp, this maximum sum would be the maximum possible sum with left and right as boundary columns. To get the overall maximum sum, we compare this sum with the maximum sum so far.
Implementation:
C++
// Program to find maximum sum subarray// in a given 2D array#include <bits/stdc++.h>using namespace std;#define ROW 4#define COL 5// Implementation of Kadane's algorithm for// 1D array. The function returns the maximum// sum and stores starting and ending indexes// of the maximum sum subarray at addresses// pointed by start and finish pointers// respectively.int kadane(int* arr, int* start, int* finish, int n){ // initialize sum, maxSum and int sum = 0, maxSum = INT_MIN, i; // Just some initial value to check // for all negative values case *finish = -1; // local variable int local_start = 0; for (i = 0; i < n; ++i) { sum += arr[i]; if (sum < 0) { sum = 0; local_start = i + 1; } else if (sum > maxSum) { maxSum = sum; *start = local_start; *finish = i; } } // There is at-least one // non-negative number if (*finish != -1) return maxSum; // Special Case: When all numbers // in arr[] are negative maxSum = arr[0]; *start = *finish = 0; // Find the maximum element in array for (i = 1; i < n; i++) { if (arr[i] > maxSum) { maxSum = arr[i]; *start = *finish = i; } } return maxSum;}// The main function that finds// maximum sum rectangle in M[][]void findMaxSum(int M[][COL]){ // Variables to store the final output int maxSum = INT_MIN, finalLeft, finalRight, finalTop, finalBottom; int left, right, i; int temp[ROW], sum, start, finish; // Set the left column for (left = 0; left < COL; ++left) { // Initialize all elements of temp as 0 memset(temp, 0, sizeof(temp)); // Set the right column for the left // column set by outer loop for (right = left; right < COL; ++right) { // Calculate sum between current left // and right for every row 'i' for (i = 0; i < ROW; ++i) temp[i] += M[i][right]; // Find the maximum sum subarray in temp[]. // The kadane() function also sets values // of start and finish. So 'sum' is sum of // rectangle between (start, left) and // (finish, right) which is the maximum sum // with boundary columns strictly as left // and right. sum = kadane(temp, &start, &finish, ROW); // Compare sum with maximum sum so far. // If sum is more, then update maxSum and // other output values if (sum > maxSum) { maxSum = sum; finalLeft = left; finalRight = right; finalTop = start; finalBottom = finish; } } } // Print final values cout << "(Top, Left) (" << finalTop << ", " << finalLeft << ")" << endl; cout << "(Bottom, Right) (" << finalBottom << ", " << finalRight << ")" << endl; cout << "Max sum is: " << maxSum << endl;}// Driver Codeint main(){ int M[ROW][COL] = { { 1, 2, -1, -4, -20 }, { -8, -3, 4, 2, 1 }, { 3, 8, 10, 1, 3 }, { -4, -1, 1, 7, -6 } }; // Function call findMaxSum(M); return 0;}// This code is contributed by// rathbhupendra |
C
// Program to find maximum sum subarray// in a given 2D array#include <limits.h>#include <stdio.h>#include <string.h>#define ROW 4#define COL 5// Implementation of Kadane's algorithm// for 1D array. The function returns the// maximum sum and stores starting and// ending indexes of the maximum sum subarray// at addresses pointed by start and finish// pointers respectively.int kadane(int* arr, int* start, int* finish, int n){ // initialize sum, maxSum and int sum = 0, maxSum = INT_MIN, i; // Just some initial value to check for all negative // values case *finish = -1; // local variable int local_start = 0; for (i = 0; i < n; ++i) { sum += arr[i]; if (sum < 0) { sum = 0; local_start = i + 1; } else if (sum > maxSum) { maxSum = sum; *start = local_start; *finish = i; } } // There is at-least one non-negative number if (*finish != -1) return maxSum; // Special Case: When all numbers in arr[] // are negative maxSum = arr[0]; *start = *finish = 0; // Find the maximum element in array for (i = 1; i < n; i++) { if (arr[i] > maxSum) { maxSum = arr[i]; *start = *finish = i; } } return maxSum;}// The main function that finds maximum// sum rectangle in// M[][]void findMaxSum(int M[][COL]){ // Variables to store the final output int maxSum = INT_MIN, finalLeft, finalRight, finalTop, finalBottom; int left, right, i; int temp[ROW], sum, start, finish; // Set the left column for (left = 0; left < COL; ++left) { // Initialize all elements of temp as 0 memset(temp, 0, sizeof(temp)); // Set the right column for the left column set by // outer loop for (right = left; right < COL; ++right) { // Calculate sum between current left and right // for every row 'i' for (i = 0; i < ROW; ++i) temp[i] += M[i][right]; // Find the maximum sum subarray in temp[]. // The kadane() function also sets values of // start and finish. So 'sum' is sum of // rectangle between (start, left) and (finish, // right) which is the maximum sum with // boundary columns strictly as left and right. sum = kadane(temp, &start, &finish, ROW); // Compare sum with maximum sum so far. If sum // is more, then update maxSum and other output // values if (sum > maxSum) { maxSum = sum; finalLeft = left; finalRight = right; finalTop = start; finalBottom = finish; } } } // Print final values printf("(Top, Left) (%d, %d)\n", finalTop, finalLeft); printf("(Bottom, Right) (%d, %d)\n", finalBottom, finalRight); printf("Max sum is: %d\n", maxSum);}// Driver Codeint main(){ int M[ROW][COL] = { { 1, 2, -1, -4, -20 }, { -8, -3, 4, 2, 1 }, { 3, 8, 10, 1, 3 }, { -4, -1, 1, 7, -6 } }; // Function call findMaxSum(M); return 0;} |
Java
// Java Program to find max sum rectangular submatriximport java.io.*;import java.lang.*;import java.util.*;class MaximumSumRectangle { // Function to find maximum sum rectangular // submatrix private static int maxSumRectangle(int[][] mat) { int m = mat.length; int n = mat[0].length; int preSum[][] = new int[m + 1][n]; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { preSum[i + 1][j] = preSum[i][j] + mat[i][j]; } } int maxSum = -1; int minSum = Integer.MIN_VALUE; int negRow = 0, negCol = 0; int rStart = 0, rEnd = 0, cStart = 0, cEnd = 0; for (int rowStart = 0; rowStart < m; rowStart++) { for (int row = rowStart; row < m; row++) { int sum = 0; int curColStart = 0; for (int col = 0; col < n; col++) { sum += preSum[row + 1][col] - preSum[rowStart][col]; if (sum < 0) { if (minSum < sum) { minSum = sum; negRow = row; negCol = col; } sum = 0; curColStart = col + 1; } else if (maxSum < sum) { maxSum = sum; rStart = rowStart; rEnd = row; cStart = curColStart; cEnd = col; } } } } // Printing final values if (maxSum == -1) { System.out.println("from row - " + negRow + " to row - " + negRow); System.out.println("from col - " + negCol + " to col - " + negCol); } else { System.out.println("from row - " + rStart + " to row - " + rEnd); System.out.println("from col - " + cStart + " to col - " + cEnd); } return maxSum == -1 ? minSum : maxSum; } // Driver Code public static void main(String[] args) { int arr[][] = new int[][] { { 1, 2, -1, -4, -20 }, { -8, -3, 4, 2, 1 }, { 3, 8, 10, 1, 3 }, { -4, -1, 1, 7, -6 } }; // Function call System.out.println(maxSumRectangle(arr)); }}// This code is contributed by Nayanava De |
Python3
# Python3 program to find maximum sum# subarray in a given 2D array# Implementation of Kadane's algorithm# for 1D array. The function returns the# maximum sum and stores starting and# ending indexes of the maximum sum subarray# at addresses pointed by start and finish# pointers respectively.def kadane(arr, start, finish, n): # initialize sum, maxSum and Sum = 0 maxSum = -999999999999 i = None # Just some initial value to check # for all negative values case finish[0] = -1 # local variable local_start = 0 for i in range(n): Sum += arr[i] if Sum < 0: Sum = 0 local_start = i + 1 elif Sum > maxSum: maxSum = Sum start[0] = local_start finish[0] = i # There is at-least one # non-negative number if finish[0] != -1: return maxSum # Special Case: When all numbers # in arr[] are negative maxSum = arr[0] start[0] = finish[0] = 0 # Find the maximum element in array for i in range(1, n): if arr[i] > maxSum: maxSum = arr[i] start[0] = finish[0] = i return maxSum# The main function that finds maximum# sum rectangle in M[][]def findMaxSum(M): global ROW, COL # Variables to store the final output maxSum, finalLeft = -999999999999, None finalRight, finalTop, finalBottom = None, None, None left, right, i = None, None, None temp = [None] * ROW Sum = 0 start = [0] finish = [0] # Set the left column for left in range(COL): # Initialize all elements of temp as 0 temp = [0] * ROW # Set the right column for the left # column set by outer loop for right in range(left, COL): # Calculate sum between current left # and right for every row 'i' for i in range(ROW): temp[i] += M[i][right] # Find the maximum sum subarray in # temp[]. The kadane() function also # sets values of start and finish. # So 'sum' is sum of rectangle between # (start, left) and (finish, right) which # is the maximum sum with boundary columns # strictly as left and right. Sum = kadane(temp, start, finish, ROW) # Compare sum with maximum sum so far. # If sum is more, then update maxSum # and other output values if Sum > maxSum: maxSum = Sum finalLeft = left finalRight = right finalTop = start[0] finalBottom = finish[0] # Prfinal values print("(Top, Left)", "(", finalTop, finalLeft, ")") print("(Bottom, Right)", "(", finalBottom, finalRight, ")") print("Max sum is:", maxSum)# Driver CodeROW = 4COL = 5M = [[1, 2, -1, -4, -20], [-8, -3, 4, 2, 1], [3, 8, 10, 1, 3], [-4, -1, 1, 7, -6]]# Function callfindMaxSum(M)# This code is contributed by PranchalK |
C#
// C# Given a 2D array, find the// maximum sum subarray in itusing System;class GFG { /** * To find maxSum in 1d array * * return {maxSum, left, right} */ public static int[] kadane(int[] a) { int[] result = new int[] { int.MinValue, 0, -1 }; int currentSum = 0; int localStart = 0; for (int i = 0; i < a.Length; i++) { currentSum += a[i]; if (currentSum < 0) { currentSum = 0; localStart = i + 1; } else if (currentSum > result[0]) { result[0] = currentSum; result[1] = localStart; result[2] = i; } } // all numbers in a are negative if (result[2] == -1) { result[0] = 0; for (int i = 0; i < a.Length; i++) { if (a[i] > result[0]) { result[0] = a[i]; result[1] = i; result[2] = i; } } } return result; } /** * To find and print maxSum, (left, top),(right, bottom) */ public static void findMaxSubMatrix(int[, ] a) { int cols = a.GetLength(1); int rows = a.GetLength(0); int[] currentResult; int maxSum = int.MinValue; int left = 0; int top = 0; int right = 0; int bottom = 0; for (int leftCol = 0; leftCol < cols; leftCol++) { int[] tmp = new int[rows]; for (int rightCol = leftCol; rightCol < cols; rightCol++) { for (int i = 0; i < rows; i++) { tmp[i] += a[i, rightCol]; } currentResult = kadane(tmp); if (currentResult[0] > maxSum) { maxSum = currentResult[0]; left = leftCol; top = currentResult[1]; right = rightCol; bottom = currentResult[2]; } } } Console.Write("MaxSum: " + maxSum + ", range: [(" + left + ", " + top + ")(" + right + ", " + bottom + ")]"); } // Driver Code public static void Main() { int[, ] arr = { { 1, 2, -1, -4, -20 }, { -8, -3, 4, 2, 1 }, { 3, 8, 10, 1, 3 }, { -4, -1, 1, 7, -6 } }; // Function call findMaxSubMatrix(arr); }}// This code is contributed// by PrinciRaj1992 |
Javascript
<script>// Program to find maximum sum subarray// in a given 2D arrayvar ROW = 4var COL = 5var start = 0var finish = 0// Implementation of Kadane's algorithm for// 1D array. The function returns the maximum// sum and stores starting and ending indexes// of the maximum sum subarray at addresses// pointed by start and finish pointers// respectively.function kadane(arr, n){ // initialize sum, maxSum and var sum = 0, maxSum = -1000000000, i; // Just some initial value to check // for all negative values case finish = -1; // local variable var local_start = 0; for (i = 0; i < n; ++i) { sum += arr[i]; if (sum < 0) { sum = 0; local_start = i + 1; } else if (sum > maxSum) { maxSum = sum; start = local_start; finish = i; } } // There is at-least one // non-negative number if (finish != -1) return maxSum; // Special Case: When all numbers // in arr[] are negative maxSum = arr[0]; start = finish = 0; // Find the maximum element in array for (i = 1; i < n; i++) { if (arr[i] > maxSum) { maxSum = arr[i]; start = finish = i; } } return maxSum;}// The main function that finds// maximum sum rectangle in M[][]function findMaxSum(M){ // Variables to store the final output var maxSum = -1000000000, finalLeft=0, finalRight=0, finalTop=0, finalBottom=0; var left, right, i; var temp = Array(ROW); var sum; // Set the left column for (left = 0; left < COL; ++left) { // Initialize all elements of temp as 0 temp = Array(ROW).fill(0); // Set the right column for the left // column set by outer loop for (right = left; right < COL; ++right) { // Calculate sum between current left // and right for every row 'i' for (i = 0; i < ROW; ++i) temp[i] += M[i][right]; // Find the maximum sum subarray in temp[]. // The kadane() function also sets values // of start and finish. So 'sum' is sum of // rectangle between (start, left) and // (finish, right) which is the maximum sum // with boundary columns strictly as left // and right. sum = kadane(temp, ROW); // Compare sum with maximum sum so far. // If sum is more, then update maxSum and // other output values if (sum > maxSum) { maxSum = sum; finalLeft = left; finalRight = right; finalTop = start; finalBottom = finish; } } } // Print final values document.write("(Top, Left) (" + finalTop + ", " + finalLeft + ")" + "<br>"); document.write("(Bottom, Right) (" + finalBottom + ", " + finalRight + ")" + "<br>"); document.write("Max sum is: " + maxSum + "<br>");}// Driver Codevar M = [ [ 1, 2, -1, -4, -20 ], [ -8, -3, 4, 2, 1 ], [ 3, 8, 10, 1, 3 ], [ -4, -1, 1, 7, -6 ] ];// Function callfindMaxSum(M);// This code is contributed by rutvik_56.</script> |
Output
(Top, Left) (1, 1) (Bottom, Right) (3, 3) Max sum is: 29
Time Complexity: O(c*c*r), where c represents the number of columns and r represents the number of rows in the given matrix.
Auxiliary Space: O(r), where r represents the number of rows in the given matrix.
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