A problem that is simple to solve in one dimension is often much more difficult to solve in more than one dimension. Consider satisfying a boolean expression in conjunctive normal form in which each conjunct consists of exactly 3 disjuncts. This problem (3-SAT) is NP-complete. The problem 2-SAT is solved quite efficiently, however. In contrast, some problems belong to the same complexity class regardless of the dimensionality of the problem.
Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle. A sub-rectangle is any contiguous sub-array of size or greater located within the whole array. As an example, the maximal sub-rectangle of the array:
is in the lower-left-hand corner:
and has the sum of 15.
The input consists of an array of integers. The input begins with a single positive integer N on a line by itself indicating the size of the square two dimensional array. This is followed by integers separated by white-space (newlines and spaces). These integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on the second row, left-to-right, etc.). N may be as large as 100. The numbers in the array will be in the range [-127, 127].
The output is the sum of the maximal sub-rectangle.
Sample Input
4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2
Sample Output
15
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace MaxSum
{
class Program
{
static void Main(string[] args)
{
//Array initialization
int[,] matrix = new int[,] {
{0, -2, -7, 0},
{9, 2, -6, 2},
{-4, 1, -4, 1},
{-1, 8, 0, -2}
};
int rows = matrix.GetLength(0);
int cols = matrix.GetLength(1);
int flag = 0, sum = 0, row = 0, col = 0, rowSize = 0, colSize = 0;
for(int i = 0; i < rows; i++)
{
for(int j = 0; j < cols; j++)
{
//i and j will determine the varying window size here
for (int p = 0; p < rows; p++)
{
for (int q = 0; q < cols; q++)
{
//p and q are the matrix elements
if (p + i < rows && q + j < cols)
{
//Call the sum function here
if (flag == 0)
{
sum = Sum(matrix, p, q, i, j);
rowSize = i;
colSize = j;
row = p;
col = q;
flag = 1;
}
else
{
if (sum < Sum(matrix, p, q, i, j))
{
sum = Sum(matrix, p, q, i, j);
rowSize = i;
colSize = j;
row = p;
col = q;
}
}
}
}
}
}
}
Console.WriteLine("Max Sum = " + sum);
Console.WriteLine("Maximal Sub Matrix is:");
for (int i = row; i <= row + rowSize; i++)
{
for (int j = col; j <= col + colSize; j++)
{
Console.Write(matrix[i, j]);
Console.Write(" ");
}
Console.WriteLine();
}
}
public static int Sum(int[,] matrix, int row, int col, int rowSize, int colSize)
{
int sum = 0;
for (int i = row; i <= row + rowSize; i++)
{
for (int j = col; j <= col + colSize; j++)
{
sum = sum + matrix[i, j];
}
}
return sum;
}
}
}
Max Sum = 15
Maximal Sub Matrix is:
9 2
-4 1
-1 8
Press any key to continue . . .