# Maximum sum subrectangle

### Finding the maximum sum subrectangle in a two dimensional array/matrix.

A brute-force way of finding the maximum sum sub-rectangle is to set the postion of the top-left and bottom-right corners of the sub-rectangle and adding the integers within it while iterating through all the rows sequentially. Setting all the top-left and bottom-right corners of the sub-rectangles within a given rectangle takes O ( n 4 ) time, adding the elements within it takes O ( n 2 ) time. Thus the simplistic brute-force algorithm takes O ( n 6 ) time.

A better ( not the best ) approach than the above is shown below that has the time complexity of O ( n4 ).

Dynamic programming approach

• Convert the given rectangle into an aggregate rectangle. In an aggregate rectangle, each integer at position r (row) and c (column) indicates the sum of a sub-rectangle whose top-left corner is ( 0, 0 ) and bottom-right corner is ( r, c ).
The aggregate rectangle can be calculated in O ( n 2 ) time and is based on the inclusion-exclusion principal.

Inclusion-Exclusion principle
1. For each row r in ROWS :
2.      For each column c in COLS :
3.          If r > 0 ( Add integers from the row above ) :
rectangle [ r ] [ c ] += rectangle [ r - 1 ] [ c ]
4.          If c > 0 ( Add integers from the column to the left ) :
rectangle [ r ] [ c ] += rectangle [ r ] [ c - 1 ]
5.          If r > 0 and c > 0
( Subtract the integer at the diagonal position that got added twice during the previous row and column additions ) :
rectangle [ r ] [ c ] -= rectangle [ r - 1 ] [ c - 1 ]

Once an aggregate rectangle as shown below is obtained, finding the maximum sum subrectangle can be done in O ( n4 ). All possible subrectangles can be obtained by setting the position of the top-left and bottom-right corner(s). The sum of the subrectangle thus formed can be deduced by using the inclusion-exclusion principle.

Example of finding the sum of a subrectangle with co-ordinates : Top-Left ( 2, 2 ) and Bottom-Right ( 3, 3 )
Left is 2 and 2 > 0 : sum of subrectangle = 2 - (-20) = 22 ( subtract the aggregate rectangle on the left )
Top is 2 and 2 > 0 : sum of subrectangle = 22 - 8 = 14 ( subtract the aggregate rectangle above )
Left and Top both are greater than 0 : sum of rectangle = 14 + 11 = 25 ( add the aggregate rectangle in the diagonal location above as it got subtracted twice )

Time complexity : O ( n4 ). If the length is ’l’ and the breadth is ‘b’ then the time complexity is O ( l2 . b2 ) which is effectively O( n4 ) if l equals b equals n.

The time complexity can be further reduced to O ( n3 ) using dynamic programming with Kadane’s algorithm.

Program for finding the maximum sum subrectangle.

``````#!/usr/bin/python3
from typing import List # For annotations

def GetMaxSumRectangle (matrix : List[List[int]]):

rows = len(matrix)
cols = len(matrix)

# Inclusion Exclusion Principle
for r in range(rows):
for c in range(cols):

# Add element from the previous column
if(r > 0):
matrix[r][c] += matrix[r-1][c]

# Add element from the previous row
if(c > 0):
matrix[r][c] += matrix[r][c-1]

# Subtract diagonal element that got added twice from previous row and column additions
if(r > 0 and c > 0):
matrix[r][c] -= matrix[r-1][c-1]

maxsum = -100*100*100

for i in range(rows):
for j in range(cols):
for k in range(i, rows):
for l in range(j, cols):

sum_subrectangle = matrix[k][l]

if(i > 0):
sum_subrectangle -= matrix[i-1][l]
if(j > 0):
sum_subrectangle -= matrix[k][j-1]
if(i > 0 and j > 0):
sum_subrectangle += matrix[i-1][j-1]

if(sum_subrectangle > maxsum):
# Co-ordinates of maximum sum sub rectangle
from_row = i
from_column = j
to_row = k
to_column = l
maxsum = sum_subrectangle

print("Maximum sum in sub-rectangle: " + str(maxsum))
print("Maximum sum rectangle co-ordinates: (" + str(from_row) + "," + str(from_column) + ")" + "-" + "(" + str(to_row) + "," + str(to_column) + ")")

matrix1 = [[  1, -2, -6, 0 ],
[  9,  3, -5, 3 ],
[ -5,  1, -4, 1 ],
[ -3,  7,  0,-3 ],
]
GetMaxSumRectangle(matrix1)

matrix2 = [[ 8, -2, -6, 11 ],
[ 9, -4, -5, -3 ],
[-5, -16, 14,-1 ],
[-3, -7, -1, 13 ],
]
GetMaxSumRectangle(matrix2)
``````

Output

``````Maximum sum in sub-rectangle: 12
Maximum sum rectangle co-ordinates: (1,0)-(1,1)
Maximum sum in sub-rectangle: 25
Maximum sum rectangle co-ordinates: (2,2)-(3,3)
``````
``````#include<iostream>
#include<vector>

using namespace std;

void GetMaxSumRectangle (vector<vector<int>> & matrix) {

int rows = matrix.size();
int cols = matrix.size();

// Inclusion Exclusion Principle
for(int r=0; r<rows; r++) {
for(int c=0; c<cols; c++) {

// Add element from the previous column
if(r > 0)
matrix[r][c] += matrix[r-1][c];
// Add element from the previous row
if(c > 0)
matrix[r][c] += matrix[r][c-1];
// Subtract diagonal element that got added twice from previous row and column additions
if(r > 0 && c > 0)
matrix[r][c] -= matrix[r-1][c-1];
}
}

int maxsum = -127*100*100;

// Co-ordinates of maximum sum sub rectangle
int from_row, from_column;
int to_row, to_column;

for(int i=0; i<rows; i++) for(int j=0; j<cols; j++)
for(int k=i; k<rows; k++) for(int l=j; l<cols; l++) {

int sum_subrectangle = matrix[k][l];

if(i > 0)
sum_subrectangle -= matrix[i-1][l];
if(j > 0)
sum_subrectangle -= matrix[k][j-1];
if(i > 0 && j > 0)
sum_subrectangle += matrix[i-1][j-1];

if(sum_subrectangle > maxsum){
from_row = i, from_column = j, to_row = k, to_column = l;
maxsum = sum_subrectangle;
}
}

cout << "Maximum sum in sub-rectangle: " << maxsum << endl;
cout << "Maximum sum rectangle co-ordinates: (" << from_row << "," << from_column << ")" << "-" << "(" << to_row << "," << to_column << ")" << endl;
}

int main(){

vector<vector<int>> matrix1 = {{  1, -2, -6, 0 },
{  9,  3, -5, 3 },
{ -5,  1, -4, 1 },
{ -3,  7,  0,-3 },
};
GetMaxSumRectangle(matrix1);

vector<vector<int>> matrix2 = {{ 8, -2, -6, 11 },
{ 9, -4, -5, -3 },
{-5, -16, 14,-1 },
{-3, -7, -1, 13 },
};
GetMaxSumRectangle(matrix2);
return 0;
};
``````

Output

``````Maximum sum in sub-rectangle: 12
Maximum sum rectangle co-ordinates: (1,0)-(1,1)
Maximum sum in sub-rectangle: 25
Maximum sum rectangle co-ordinates: (2,2)-(3,3)
``````