# Maximum sum rectangle in a 2D matrix | DP-27

Given a 2D array, find the maximum sum subarray in it. For example, in the following 2D array, the maximum sum subarray is highlighted with blue rectangle and sum of this subarray is 29. This problem is mainly an extension of Largest Sum Contiguous Subarray for 1D array.

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

The naive solution for this problem is to check every possible rectangle in given 2D array. This solution requires 4 nested loops and time complexity of this solution would be O(n^4).

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 sun 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.

## C++

 `// Program to find maximum sum subarray  ` `// in a given 2D array  ` `#include  ` `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;  ` `    ``*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 Code ` `int` `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}};  ` ` `  `    ``findMaxSum(M);  ` ` `  `    ``return` `0;  ` `}  ` ` `  `// This code is contributed by ` `// rathbhupendra `

## C

 `// Program to find maximum sum subarray in a given 2D array ` `#include ` `#include ` `#include ` `#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; ` `    ``*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 program to test above functions ` `int` `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} ` `                      ``}; ` ` `  `    ``findMaxSum(M); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java Program to find max sum rectangular submatrix ` ` `  `import` `java.util.*; ` `import` `java.lang.*; ` `import` `java.io.*; ` `  `  `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``}}; ` `        ``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 Code ` `ROW ``=` `4` `COL ``=` `5` `M ``=` `[[``1``, ``2``, ``-``1``, ``-``4``, ``-``20``], ` `     ``[``-``8``, ``-``3``, ``4``, ``2``, ``1``],  ` `     ``[``3``, ``8``, ``10``, ``1``, ``3``],  ` `     ``[``-``4``, ``-``1``, ``1``, ``7``, ``-``6``]]  ` ` `  `findMaxSum(M) ` ` `  `# This code is contributed by PranchalK `

## C#

 `// C# Given a 2D array, find the  ` `// maximum sum subarray in it  ` `using` `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)  ` `        ``{  ` `            ``result = currentSum;  ` `            ``result = localStart;  ` `            ``result = i;  ` `        ``}  ` `    ``}  ` `     `  `    ``// all numbers in a are negative  ` `    ``if` `(result == -1)  ` `    ``{  ` `        ``result = 0;  ` `        ``for` `(``int` `i = 0; i < a.Length; i++)  ` `        ``{  ` `            ``if` `(a[i] > result)  ` `            ``{  ` `                ``result = a[i];  ` `                ``result = i;  ` `                ``result = 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 > maxSum)  ` `            ``{  ` `                ``maxSum = currentResult;  ` `                ``left = leftCol;  ` `                ``top = currentResult;  ` `                ``right = rightCol;  ` `                ``bottom = currentResult;  ` `            ``}  ` `        ``}  ` `    ``}  ` `     `  `    ``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} };  ` `    ``findMaxSubMatrix(arr); ` `}  ` `}  ` ` `  `// This code is contributed  ` `// by PrinciRaj1992 `

Output:

```(Top, Left) (1, 1)
(Bottom, Right) (3, 3)
Max sum is: 29```

Time Complexity: O(n^3)

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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