zoukankan      html  css  js  c++  java
  • [Coding Made Simple] Maximum Sum Rectangular Submatrix in Matrix

    Given a 2D matrix of  integers, find maximum sum rectangle in this matrix.

    Brute force solution of enumerating all possible submatrix then check their sum is too inefficient. 

    Algorithm of using the Kadane's algorithm(Maximum Subarray)

    1. Fix a left and right column range to get a submatrix, there are O(col * col) such submatrices to consider.

    2. For each submatrix, use a 1D array to get its total sum in O(row) time(dynamic programming used to avoid starting each computation from scratch).

        Then apply the Kadane's algorithm to get the maximum subarray sum on this 1D array in O(row) time. Because this 1D array represents the total

      sum of the current submatrix, the max subarray sum from this 1D array represents the max rectangular submatrix sum of the current submatrix.

    O(col * col * row) runtime, O(row) space

      1 import java.util.ArrayList;
      2 
      3 class Point {
      4     int row, col;
      5     Point(int r, int c) {
      6         row = r;
      7         col = c;
      8     }
      9     void display() {
     10         System.out.println(this.row + ", " + this.col);
     11     }
     12 }
     13 public class MaxRectangleSum {
     14     private ArrayList<Point> maxRecCoordinates = null;
     15     public int getMaxRecSum(int[][] matrix) {
     16         if(matrix == null || matrix.length == 0 || matrix[0].length == 0) {
     17             return 0;
     18         }
     19         maxRecCoordinates = new ArrayList<Point>();
     20         int max = Integer.MIN_VALUE, x = 0, y = 0;
     21         boolean hasPositiveInt = false;
     22         
     23         //preprocess matrix for case of no positive integer present
     24         for(int i = 0; i < matrix.length; i++) {
     25             for(int j = 0; j < matrix[0].length; j++) {
     26                 if(matrix[i][j] > 0) {
     27                     hasPositiveInt = true;
     28                     break;
     29                 }
     30                 else if(matrix[i][j] > max){
     31                     max = matrix[i][j];
     32                     x = i;
     33                     y = j;
     34                 }
     35             }
     36         }
     37         if(!hasPositiveInt) {
     38             for(int i = 0; i < 2; i++) {
     39                 maxRecCoordinates.add(new Point(x, y));
     40             }
     41             return max;
     42         }
     43         
     44         max = Integer.MIN_VALUE;
     45         int[] range= new int[2];
     46         int maxLeft = 0, maxRight = 0, maxTop = 0, maxBottom = 0;
     47         int[] colArr = new int[matrix.length];
     48         for(int l = 0; l < matrix[0].length; l++) {
     49             for(int i = 0; i < matrix.length; i++) {
     50                 colArr[i] = 0;
     51             }
     52             for(int r = l; r < matrix[0].length; r++) {
     53                 for(int i = 0; i < matrix.length; i++) {
     54                     colArr[i] += matrix[i][r]; 
     55                 }
     56                 int currSum = getMaxSubarraySum(colArr, range);
     57                 if(currSum > max) {
     58                     max = currSum;
     59                     maxLeft = l;
     60                     maxRight = r;
     61                     maxTop = range[0];
     62                     maxBottom = range[1];
     63                 }
     64             }
     65         }
     66         Point topLeft = new Point(maxTop, maxLeft);
     67         Point bottomRight = new Point(maxBottom, maxRight);
     68         maxRecCoordinates.add(topLeft);
     69         maxRecCoordinates.add(bottomRight);
     70         return max;
     71     }
     72     private int getMaxSubarraySum(int[] arr, int[] range) {
     73         range[0] = 0;
     74         range[1] = 0;
     75         int currSum = 0, minSum = 0, maxSum = Integer.MIN_VALUE;
     76         int minSumEndIdx = -1;
     77         for(int i = 0; i < arr.length; i++) {
     78             currSum += arr[i];
     79             if(currSum - minSum > maxSum) {
     80                 maxSum = currSum - minSum;
     81                 range[0] = minSumEndIdx + 1;
     82                 range[1] = i;
     83             }
     84             if(currSum < minSum) {
     85                 minSum = currSum;
     86                 minSumEndIdx = i;
     87             }
     88         }
     89         return maxSum;
     90     }
     91     public static void main(String[] args) {
     92         int[][] matrix1 = {{2,1,-3,-4,5}, {0,6,3,4,1},{2,-2,-1,4,-5},{-3,3,1,0,3}};
     93         int[] arr = {-2,2,-3,4,-1,2,1,-5,3};
     94         int[] range = new int[2];
     95         MaxRectangleSum test = new MaxRectangleSum();
     96         System.out.println(test.getMaxSubarraySum(arr, range));
     97         for(int i = 0; i < 2; i++) {
     98             System.out.println(range[i]);
     99         }
    100         System.out.println(test.getMaxRecSum(matrix1));
    101         for(int i = 0; i < test.maxRecCoordinates.size(); i++) {
    102             test.maxRecCoordinates.get(i).display();
    103         }
    104     }
    105 }

    Related Problems 

    [LintCode] Maximum Subarray

    Submatrix Sum To Zero

  • 相关阅读:
    django DEBUG=False
    Lftp 简单使用步骤
    django admin管理后台中文添加问题
    Rsync同步设置的一例
    在nginx中,禁止IP访问.只可以使用域名访问.
    python imaplib无痕取信的主要
    Centos安装配置Postfix邮件服务器--网址
    还没有写完准备弡上cpickle 还有字典
    python 截取某一天的日志,简单操作
    abp 关闭审计日志
  • 原文地址:https://www.cnblogs.com/lz87/p/7288814.html
Copyright © 2011-2022 走看看