Skip to content
Related Articles

Related Articles

Improve Article
Java Program for Maximum sum rectangle in a 2D matrix | DP-27
  • Last Updated : 12 Dec, 2018

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.

Java




import java.util.*;
import java.lang.*;
import java.io.*;
  
/**
 * Given a 2D array, find the maximum sum subarray in it
 */
class Ideone {
    public static void main(String[] args) throws java.lang.Exception
    {
        findMaxSubMatrix(new int[][] {
            { 1, 2, -1, -4, -20 },
            { -8, -3, 4, 2, 1 },
            { 3, 8, 10, 1, 3 },
            { -4, -1, 1, 7, -6 } });
    }
  
    /**
     * To find maxSum in 1d array
     
     * return {maxSum, left, right}
     */
    public static int[] kadane(int[] a)
    {
        // result[0] == maxSum, result[1] == start, result[2] == end;
        int[] result = new int[] { Integer.MIN_VALUE, 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[0].length;
        int rows = a.length;
        int[] currentResult;
        int maxSum = Integer.MIN_VALUE;
        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];
                }
            }
        }
        System.out.println("MaxSum: " + maxSum + ", 
         range: [(" + left + ", " + top + ")(" + right + ", " + bottom + ")]");
    }
}
// Thanks to Ilia Savin for contributing this code.
Output:
MaxSum: 29, range: [(1, 1)(3, 3)]

Please refer complete article on Maximum sum rectangle in a 2D matrix | DP-27 for more details!




My Personal Notes arrow_drop_up
Recommended Articles
Page :